\displaystyle \textbf{Question 1: }\ \text{Find the points of local maxima or local minima, if any, of the following} \\ \text{functions. Find also the local maximum or local minimum values, as the case may be:}
\displaystyle \ f(x)=\sin x-\cos x,\ \text{where}\ 0<x<2\pi\ \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \ \text{We have,}
\displaystyle f(x)=\sin x-\cos x,\ \text{where}\ 0<x<2\pi
\displaystyle f'(x)=\cos x+\sin x
\displaystyle \text{At points of local maximum and local minimum, we must have}
\displaystyle f'(x)=0
\displaystyle \Rightarrow\ \cos x+\sin x=0
\displaystyle \Rightarrow\ \sin x=-\cos x\Rightarrow\ \tan x=-1\Rightarrow\ x=\frac{3\pi}{4},\ \text{or}\ x=\frac{7\pi}{4}\ [0<x<2\pi]
\displaystyle \text{Thus, }x=\frac{3\pi}{4}\text{ and }x=\frac{7\pi}{4}\text{ are possible points of local maximum or minimum}
\displaystyle \text{Now, we test the function at each of these points}
\displaystyle \text{Clearly, }f''(x)=-\sin x+\cos x
\displaystyle \text{At }x=\frac{3\pi}{4},\ \text{We have,}
\displaystyle f''\left(\frac{3\pi}{4}\right)=-\sin\frac{3\pi}{4}+\cos\frac{3\pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2}}<0
\displaystyle \text{So, }x=\frac{3\pi}{4}\text{ is the point of local maximum}
\displaystyle \text{The local maximum value is }f\left(\frac{3\pi}{4}\right)=\sin\frac{3\pi}{4}-\cos\frac{3\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}
\displaystyle \text{At }x=\frac{7\pi}{4},\ \text{We have,}
\displaystyle f''\left(\frac{7\pi}{4}\right)=-\sin\frac{7\pi}{4}+\cos\frac{7\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}>0
\displaystyle \text{So, the function attains a local minimum at }x=\frac{7\pi}{4}
\displaystyle \text{The local minimum value is }f\left(\frac{7\pi}{4}\right)=\sin\frac{7\pi}{4}-\cos\frac{7\pi}{4}=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}

\displaystyle \textbf{Question 2: }\ \text{Find the minimum value of }ax+by,\ \text{where }xy=c^{2}\ \text{and }a,b,c \\ \text{are positive.}\ \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }z=ax+by,\ \text{where }xy=c^{2}.\ \text{Then,}
\displaystyle z=ax+\frac{bc^{2}}{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)\ \ \ \ \ \ \ \ \ \ \ \ \ \left[\because\ xy=c^{2}\Rightarrow y=\frac{c^{2}}{x}\right]
\displaystyle \Rightarrow\ \frac{dz}{dx}=a-\frac{bc^{2}}{x^{2}}\ \text{and}\ \frac{d^{2}z}{dx^{2}}=\frac{2bc^{2}}{x^{3}}
\displaystyle \text{The critical points of }z\text{ are given by }\frac{dz}{dx}=0.
\displaystyle \therefore\ \frac{dz}{dx}=0
\displaystyle \Rightarrow\ a-\frac{bc^{2}}{x^{2}}=0\Rightarrow x^{2}=\frac{bc^{2}}{a}\Rightarrow x=\pm\sqrt{\frac{b}{a}}\,c
\displaystyle \text{At }x=\sqrt{\frac{b}{a}}\,c:\ \text{We find that}
\displaystyle \frac{d^{2}z}{dx^{2}}=2bc^{2}\left(\sqrt{\frac{a}{b}}\times\frac{1}{c}\right)^{3}=\frac{2a}{c}\sqrt{\frac{a}{b}}>0
\displaystyle \text{So, }z\text{ is minimum at }x=c\sqrt{\frac{b}{a}}.
\displaystyle \text{The minimum value of }z\text{ is given by}
\displaystyle z=a\sqrt{\frac{b}{a}}\,c+\frac{bc^{2}}{c}\sqrt{\frac{a}{b}}=2\sqrt{ab}\,c\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[\text{Putting }x=\sqrt{\frac{b}{a}}\,c\ \text{in }(i)\right]
\displaystyle \text{At }x=-\sqrt{\frac{b}{a}}\,c:\ \text{We find that}
\displaystyle \frac{d^{2}z}{dx^{2}}=2bc^{2}\left(-\frac{a}{bc^{3}}\sqrt{\frac{a}{b}}\right)=-\frac{2a}{c}\sqrt{\frac{a}{b}}<0
\displaystyle \text{So, }z\text{ is maximum at }x=-\sqrt{\frac{b}{a}}\,c.

\displaystyle \textbf{Question 3: }\ \text{Show that of all the rectangles of given area, the square has the} \\ \text{smallest perimeter.}\ \text{[CBSE 2011]}
\displaystyle \text{Answer:}

\displaystyle  \text{Let }x\text{ and }y\text{ be the lengths of two sides of a rectangle of given area }A, \text{and let }P \\ \text{be the perimeter. Then,}
\displaystyle A=xy\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{and,}\ \ \ P=2(x+y)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \text{Now,}\ A=xy\Rightarrow y=\frac{A}{x}
\displaystyle \therefore\ P=2(x+y)=2\left(x+\frac{A}{x}\right)
\displaystyle \Rightarrow\ \frac{dP}{dx}=2\left(1-\frac{A}{x^{2}}\right)\ \text{and}\ \frac{d^{2}P}{dx^{2}}=\frac{4A}{x^{3}}
\displaystyle \text{The critical points of }P\text{ are given by }\frac{dP}{dx}=0.
\displaystyle \therefore\ \frac{dP}{dx}=0\Rightarrow 2\left(1-\frac{A}{x^{2}}\right)=0\Rightarrow 1-\frac{A}{x^{2}}=0\Rightarrow x^{2}=A\Rightarrow x^{2}=xy\Rightarrow x=y
\displaystyle \text{Clearly,}\ \frac{d^{2}P}{dx^{2}}=\frac{4A}{x^{3}}>0\ \text{for all positive values of }x.
\displaystyle \text{Hence, }P\text{ is minimum when }x=y\ \text{i.e. the rectangle is a square.}

\displaystyle \textbf{Question 4: }\ \text{Show that of all the rectangles inscribed in a given circle, the square} \\ \text{has the maximum area.}\ \text{[CBSE 2002, 2006, 2008, 2011, 2013]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }ABCD\text{ be a rectangle inscribed in a given circle with centre at }O\text{ and radius }a. \\ \text{Let }AB=2x\text{ and }BC=2y.\ \text{Applying Pythagoras theorem in right triangle} \\ OAM,\ \text{we obtain}
\displaystyle OA^{2}=AM^{2}+OM^{2}\Rightarrow a^{2}=x^{2}+y^{2}\Rightarrow y=\sqrt{a^{2}-x^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }A\text{ be the area of the rectangle }ABCD.\ \text{Then,}
\displaystyle A=4xy=4x\sqrt{a^{2}-x^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ \frac{dA}{dx}=4\left\{\sqrt{a^{2}-x^{2}}-\frac{x^{2}}{\sqrt{a^{2}-x^{2}}}\right\}=4\left\{\frac{a^{2}-2x^{2}}{\sqrt{a^{2}-x^{2}}}\right\}
\displaystyle \text{The critical points of }A\text{ are given by }\frac{dA}{dx}=0.
\displaystyle \therefore\ \frac{dA}{dx}=0
\displaystyle \Rightarrow\ 4\left\{\frac{a^{2}-2x^{2}}{\sqrt{a^{2}-x^{2}}}\right\}=0\Rightarrow a^{2}-2x^{2}=0\Rightarrow x=\frac{a}{\sqrt{2}}
\displaystyle \text{Now,}\ \frac{d^{2}A}{dx^{2}}=4\frac{d}{dx}\left[(a^{2}-2x^{2})(a^{2}-x^{2})^{-1/2}\right]
\displaystyle \Rightarrow\ \frac{d^{2}A}{dx^{2}}=4\left[-4x(a^{2}-x^{2})^{-1/2}+(a^{2}-2x^{2})\left(-\frac{1}{2}\right)(a^{2}-x^{2})^{-3/2}(-2x)\right]
\displaystyle \Rightarrow\ \frac{d^{2}A}{dx^{2}}=4\left[\frac{-4x}{\sqrt{a^{2}-x^{2}}}+\frac{x(a^{2}-2x^{2})}{(a^{2}-x^{2})^{3/2}}\right]
\displaystyle \left(\frac{d^{2}A}{dx^{2}}\right)_{x=\frac{a}{\sqrt{2}}}=-16<0
\displaystyle \text{Thus, }A\text{ is maximum when }x=\frac{a}{\sqrt{2}}.\ \text{Putting }x=\frac{a}{\sqrt{2}}\ \text{in (i), we get }y=\frac{a}{\sqrt{2}}.
\displaystyle \text{Therefore, }x=y=\frac{a}{\sqrt{2}}\Rightarrow 2x=2y=\sqrt{2}a\Rightarrow AB=BC\Rightarrow ABCD\text{ is a square.}
\displaystyle \text{Hence, area }A\text{ is maximum when the rectangle is a square.}

\displaystyle \textbf{Question 5: }\ \text{Show that the rectangle of maximum perimeter which can be inscribed} \\ \text{in a circle of radius }a\text{ is a square of side }\sqrt{2}a.\ \text{[CBSE 2002]}
\displaystyle \text{Answer:} \displaystyle \text{Let }ABCD\text{ be a rectangle in a given circle of radius }a\text{ with centre at }O.\ \text{Let }AB=2x\text{ and }AD=2y\text{ be the sides of the rectangle. Applying Pythagoras theorem in} \\ \triangle OAM,\ \text{we get}
\displaystyle AM^{2}+OM^{2}=OA^{2}\Rightarrow x^{2}+y^{2}=a^{2}\Rightarrow y=\sqrt{a^{2}-x^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }P\text{ be the perimeter of the rectangle }ABCD.\ \text{Then,}
\displaystyle P=4x+4y
\displaystyle \Rightarrow\ P=4x+4\sqrt{a^{2}-x^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ \frac{dP}{dx}=4-\frac{4x}{\sqrt{a^{2}-x^{2}}}
\displaystyle \text{The critical points of }P\text{ are given by }\frac{dP}{dx}=0.
\displaystyle \therefore\ \frac{dP}{dx}=0
\displaystyle \Rightarrow\ 4-\frac{4x}{\sqrt{a^{2}-x^{2}}}=0
\displaystyle \Rightarrow\ 4=\frac{4x}{\sqrt{a^{2}-x^{2}}}\Rightarrow \sqrt{a^{2}-x^{2}}=x\Rightarrow a^{2}-x^{2}=x^{2}\Rightarrow 2x^{2}=a^{2}\Rightarrow x=\frac{a}{\sqrt{2}}
\displaystyle \text{Now,}\ \frac{dP}{dx}=4-\frac{4x}{\sqrt{a^{2}-x^{2}}}
\displaystyle \Rightarrow\ \frac{d^{2}P}{dx^{2}}=-4\left\{\frac{\sqrt{a^{2}-x^{2}}-\frac{x(-x)}{\sqrt{a^{2}-x^{2}}}}{a^{2}-x^{2}}\right\}
\displaystyle \Rightarrow\ \frac{d^{2}P}{dx^{2}}=-\frac{4a^{2}}{(a^{2}-x^{2})^{3/2}}
\displaystyle \left(\frac{d^{2}P}{dx^{2}}\right)_{x=\frac{a}{\sqrt{2}}}=\frac{-4a^{2}}{\left(a^{2}-\frac{a^{2}}{2}\right)^{3/2}}=\frac{-8\sqrt{2}}{a}<0
\displaystyle \text{Thus, }P\text{ is maximum when }x=\frac{a}{\sqrt{2}}.
\displaystyle \text{Putting }x=\frac{a}{\sqrt{2}}\text{ in (i), we obtain }y=\frac{a}{\sqrt{2}}.
\displaystyle \therefore\ x=y=\frac{a}{\sqrt{2}}\Rightarrow 2x=2y\Rightarrow AB=BC\Rightarrow ABCD\text{ is a square.}
\displaystyle \text{Hence, }P\text{ is maximum when the rectangle is square of side }2x=\frac{2a}{\sqrt{2}}=\sqrt{2}a.

\displaystyle \textbf{Question 6: }\ \text{Tangent to the circle }x^{2}+y^{2}=a^{2}\ \text{at any point on it}\text{in the} \\ \text{first quadrant makes intercepts }OA\text{ and }OB\text{ on }x\text{ and }y\text{axes respectively, } \\ O\text{ being the centre of the circle. Find the minimum value of} OA+OB.\ \text{[CBSE 2015]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }P(a\cos\theta,\ a\sin\theta)\ \text{be an arbitrary point on the}
\displaystyle \text{circle }x^{2}+y^{2}=a^{2}.\ \text{If }P\text{ lies in the first quadrant, then }0\leq\theta\leq\frac{\pi}{2}.
\displaystyle \text{The equation of tangent to }x^{2}+y^{2}=a^{2}\ \text{at }P(a\cos\theta,\ a\sin\theta)\ \text{is}
\displaystyle x\cos\theta+y\sin\theta=a\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left[\text{The tangent to }x^{2}+y^{2}=a^{2}\ \text{at }(x_{1},y_{1})\ \text{is }xx_{1}+yy_{1}=a^{2}\right]
\displaystyle \text{This cuts }x\text{ and }y\text{-axis at }A(a\sec\theta,\ 0)\ \text{and }B(0,\ a\ \mathrm{cosec}\,\theta)\ \text{respectively.}
\displaystyle \therefore\ OA=a\sec\theta\ \text{and}\ OB=a\ \mathrm{cosec}\,\theta
\displaystyle \text{Let }S=OA+OB.\ \text{Then,}
\displaystyle S=a(\sec\theta+\mathrm{cosec}\,\theta)
\displaystyle \therefore\ \frac{dS}{d\theta}=a(\sec\theta\tan\theta-\mathrm{cosec}\,\theta\cot\theta)
\displaystyle \text{and,}\ \frac{d^{2}S}{d\theta^{2}}=a(\sec^{3}\theta+\sec\theta\tan^{2}\theta+\mathrm{cosec}^{3}\theta+\mathrm{cosec}\,\theta\cot^{2}\theta)
\displaystyle \text{For maximum or minimum values of }S,\ \text{we must have}
\displaystyle \frac{dS}{d\theta}=0\Rightarrow a(\sec\theta\tan\theta-\mathrm{cosec}\,\theta\cot\theta)=0\Rightarrow \tan^{3}\theta=1\Rightarrow \tan\theta=1\Rightarrow \theta=\frac{\pi}{4}
\displaystyle \text{At }\theta=\frac{\pi}{4},\ \text{we obtain}
\displaystyle \frac{d^{2}S}{d\theta^{2}}=a(2\sqrt{2}+\sqrt{2}+2\sqrt{2}+\sqrt{2})=6\sqrt{2}a>0
\displaystyle \text{Hence, }S\text{ is minimum at }\theta=\frac{\pi}{4}\ \text{and the minimum value of }S\text{ is given by}
\displaystyle S=a\left(\sec\frac{\pi}{4}+\mathrm{cosec}\,\frac{\pi}{4}\right)=2\sqrt{2}a

\displaystyle \textbf{Question 7: }\ \text{If the sum of the lengths of the hypotenuse and a side of a right}
\displaystyle \text{angled triangle is given, show that the area of the triangle is maximum when the}
\displaystyle \text{angle between them is }\frac{\pi}{3}.\ \text{[CBSE 2009, 2014, 2016, 2017]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }ABC\text{ be a right angled triangle with base }BC=x\ \text{and}
\displaystyle \text{hypotenuse }AC=y\ \text{such that }x+y=k,\ \text{where }k\ \text{is a constant. Let }\theta
\displaystyle \text{be the angle between the base and hypotenuse. Let }A\ \text{be the area of the}
\displaystyle \text{triangle. Then,}
\displaystyle A=\frac{1}{2}\ BC\times AC=\frac{1}{2}\ x\sqrt{y^{2}-x^{2}}
\displaystyle \Rightarrow\ A^{2}=\frac{x^{2}}{4}\ (y^{2}-x^{2})
\displaystyle \Rightarrow\ A^{2}=\frac{x^{2}}{4}\ \{(k-x)^{2}-x^{2}\}\ \ \ \ \ [\because\ x+y=k]
\displaystyle \Rightarrow\ A^{2}=\frac{k^{2}x^{2}-2kx^{3}}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Differentiating with respect to }x,\ \text{we get}
\displaystyle 2A\ \frac{dA}{dx}=\frac{2k^{2}x-6kx^{2}}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \Rightarrow\ \frac{dA}{dx}=\frac{k^{2}x-3kx^{2}}{4A}
\displaystyle \text{The critical numbers of }A\ \text{are given by }\frac{dA}{dx}=0.
\displaystyle \text{Now,}\ \frac{dA}{dx}=0\Rightarrow \frac{k^{2}x-3kx^{2}}{4A}=0\Rightarrow x=\frac{k}{3}.
\displaystyle \text{Differentiating (ii) with respect to }x,\ \text{we get}
\displaystyle 2\left(\frac{dA}{dx}\right)^{2}+2A\ \frac{d^{2}A}{dx^{2}}=\frac{2k^{2}-12kx}{4}\ \ \ \ ...(iii)
\displaystyle \text{When }x=\frac{k}{3},\ \frac{dA}{dx}=0.\ \text{Putting }\frac{dA}{dx}=0\ \text{and }x=\frac{k}{3}\ \text{in (iii), we get}
\displaystyle \frac{d^{2}A}{dx^{2}}=-\frac{k^{2}}{4A}<0.
\displaystyle \text{Thus, }A\ \text{is maximum when }x=\frac{k}{3}.\ \text{Putting }x=\frac{k}{3}\ \text{in }x+y=k,\ \text{we obtain }y=\frac{2k}{3}.
\displaystyle \text{In }\triangle ACB,\ \cos\theta=\frac{BC}{AB}\Rightarrow \cos\theta=\frac{x}{y}\Rightarrow \cos\theta=\frac{k/3}{2k/3}=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{3}.
\displaystyle \text{Thus, area of triangle }ABC\ \text{is maximum, when angle }\theta\ \text{between base }BC\ \text{and} \\ \text{hypotenuse }AB\ \text{is }\frac{\pi}{3}.

\displaystyle \textbf{Question 8: }\ \text{An open tank with a square base and vertical sides is to be constructed}
\displaystyle \text{from a metal sheet so as to hold a given quantity of water. Show that the cost}
\displaystyle \text{of the material will be least when depth of the tank is half of its width.} \text{[CBSE 2007, 2010]}
\displaystyle \text{Answer:} \displaystyle  \text{Let the length, width and height of the open tank be }x,\ x\ \text{and }y
\displaystyle \text{units respectively. Then, its volume is }x^{2}y\ \text{and the total surface area is }x^{2}+4xy.
\displaystyle \text{It is given that the tank can hold a given quantity of water. This means that its}
\displaystyle \text{volume is constant. Let it be }V.\ \text{Then,}
\displaystyle V=x^{2}y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{The cost of the material will be least if the total surface area is least. Let }S
\displaystyle \text{denote the total surface area. Then,}
\displaystyle S=x^{2}+4xy\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \text{We have to minimize }S\ \text{subject to the condition that the volume }V\ \text{is constant.}
\displaystyle \text{Now,}\ \ \ \ S=x^{2}+4xy
\displaystyle \Rightarrow\ S=x^{2}+\frac{4V}{x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ \frac{dS}{dx}=2x-\frac{4V}{x^{2}}\ \text{and}\ \frac{d^{2}S}{dx^{2}}=2+\frac{8V}{x^{3}}
\displaystyle \text{The critical numbers of }S\ \text{are given by }\frac{dS}{dx}=0.
\displaystyle \text{Now,}\ \frac{dS}{dx}=0
\displaystyle \Rightarrow\ 2x-\frac{4V}{x^{2}}=0
\displaystyle \Rightarrow\ 2x^{3}=4V
\displaystyle \Rightarrow\ 2x^{3}=4x^{2}y
\displaystyle \Rightarrow\ x=2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\because\ V=x^{2}y]
\displaystyle \text{Clearly,}\ \frac{d^{2}S}{dx^{2}}=2+\frac{8V}{x^{3}}>0\ \text{for all }x.
\displaystyle \text{Hence, }S\text{ is minimum when }x=2y\ \text{i.e. the depth (height) of the tank is half of its width.}

\displaystyle \textbf{Question 9: }\ \text{An open box with a square base is to be made out of a given}
\displaystyle \text{quantity of card board of area }c^{2}\ \text{square units. Show that the maximum volume}
\displaystyle \text{of the box is }\frac{c^{3}}{6\sqrt{3}}\ \text{cubic units.} \text{[CBSE 2001C, 05, 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let the length, breadth and height of the box }x,\ x\ \text{and }y
\displaystyle \text{units respectively. It is given that the area of the card board is }c^{2}\ \text{sq. units.}
\displaystyle \therefore\ x^{2}+4xy=c^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }V\text{ be the volume of the box. Then,}
\displaystyle V=x^{2}y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \Rightarrow\ V=x^{2}\left(\frac{c^{2}-x^{2}}{4x}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ V=\frac{c^{2}}{4}x-\frac{x^{3}}{4}
\displaystyle \Rightarrow\ \frac{dV}{dx}=\frac{c^{2}}{4}-\frac{3x^{2}}{4}\ \text{and}\ \frac{d^{2}V}{dx^{2}}=-\frac{3x}{2}
\displaystyle \text{The critical points of }V\text{ are given by }\frac{dV}{dx}=0.
\displaystyle \text{Now,}\ \frac{dV}{dx}=0\Rightarrow \frac{c^{2}}{4}-\frac{3x^{2}}{4}=0\Rightarrow x=\frac{c}{\sqrt{3}}.
\displaystyle \text{Clearly,}\ \left(\frac{d^{2}V}{dx^{2}}\right)_{x=\frac{c}{\sqrt{3}}}=-\frac{3c}{2\sqrt{3}}<0.
\displaystyle \text{Thus, }V\text{ is maximum when }x=\frac{c}{\sqrt{3}}.
\displaystyle \text{Putting }x=\frac{c}{\sqrt{3}}\ \text{in (i), we obtain }y=\frac{c}{2\sqrt{3}}.
\displaystyle \text{Putting }x=\frac{c}{\sqrt{3}}\ \text{and }y=\frac{c}{2\sqrt{3}}\ \text{in (ii) the maximum volume of the box is given by}
\displaystyle V=\frac{c^{2}}{3}\times\frac{c}{2\sqrt{3}}=\frac{c^{3}}{6\sqrt{3}}\ \text{cubic units.}

\displaystyle \textbf{Question 10: }\ \text{The sum of the surface areas of a rectangular parallelepiped with}
\displaystyle \text{sides }x,\ 2x\ \text{and }\frac{x}{3}\ \text{and a sphere is given to be constant. Prove that}
\displaystyle \text{the sum of the volumes is minimum, if }x\ \text{is equal to three times the radius of}
\displaystyle \text{the sphere. Also, find the minimum value of the sum of their volumes.}
\displaystyle \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }y\text{ be the radius of the sphere and let }S\text{ be the constant value of the sum of the}
\displaystyle \text{surface areas of the parallelepiped and the sphere. Then,}
\displaystyle S=2\left\{x\times 2x+2x\times\frac{x}{3}+\frac{x}{3}\times x\right\}+4\pi y^{2}
\displaystyle \text{or,}\ \ S=6x^{2}+4\pi y^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }V\text{ be the sum of the volumes of the sphere and the parallelepiped. Then,}
\displaystyle V=\frac{4}{3}\pi y^{3}+x\times 2x\times\frac{x}{3}
\displaystyle \Rightarrow\ V=\frac{4}{3}\pi y^{3}+\frac{2}{3}x^{3}
\displaystyle \Rightarrow\ V=\frac{4}{3}\pi\left(\frac{S-6x^{2}}{4\pi}\right)^{3/2}+\frac{2}{3}x^{3}\ \ \ \ \ \left[\because\ S=6x^{2}+4\pi y^{2}\Rightarrow y^{2}=\frac{S-6x^{2}}{4\pi}\right]
\displaystyle \Rightarrow\ V=\frac{1}{6\sqrt{\pi}}(S-6x^{2})^{3/2}+\frac{2}{3}x^{3}
\displaystyle \Rightarrow\ \frac{dV}{dx}=\frac{1}{6\sqrt{\pi}}\times\frac{3}{2}(S-6x^{2})^{1/2}(-12x)+\frac{2}{3}\times 3x^{2}
\displaystyle \Rightarrow\ \frac{dV}{dx}=-\frac{3}{\sqrt{\pi}}(S-6x^{2})^{1/2}x+2x^{2}\ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \text{The critical numbers of }V\text{ are given by }\frac{dV}{dx}=0.
\displaystyle \text{Now,}\ \frac{dV}{dx}=0
\displaystyle \Rightarrow\ -\frac{3}{\sqrt{\pi}}(S-6x^{2})^{1/2}x+2x^{2}=0
\displaystyle \Rightarrow\ \frac{3x}{\sqrt{\pi}}(S-6x^{2})^{1/2}=2x^{2}
\displaystyle \Rightarrow\ \frac{3}{\sqrt{\pi}}(S-6x^{2})^{1/2}=2x
\displaystyle \Rightarrow\ 9(S-6x^{2})=4\pi x^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Squaring both sides}]
\displaystyle \Rightarrow\ 9(4\pi y^{2})=4\pi x^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ 9y^{2}=x^{2}
\displaystyle \Rightarrow\ x=3y
\displaystyle \text{Putting }x=3y,\ \text{or, }y=\frac{x}{3}\ \text{in (i), we obtain }S=6x^{2}+\frac{4\pi x^{2}}{9}.
\displaystyle \text{Differentiating (ii), we obtain}
\displaystyle \frac{d^{2}V}{dx^{2}}=-\frac{3}{\sqrt{\pi}}(S-6x^{2})^{1/2}+\frac{18x^{2}}{\sqrt{\pi}\sqrt{S-6x^{2}}}+4x
\displaystyle \text{When }x=3y\ \text{or, }y=\frac{x}{3},\ \text{we obtain}
\displaystyle \frac{d^{2}V}{dx^{2}}=-\frac{3}{\sqrt{\pi}}\left(\frac{4\pi x^{2}}{9}\right)^{1/2}+\frac{18x^{2}}{\sqrt{\pi}\left(\frac{2}{3}\sqrt{\pi}x\right)}+4x=-2x+\frac{27x}{\pi}+4x=\frac{27x}{\pi}+2x>0
\displaystyle \text{So, }V\text{ is minimum when }x=3y.
\displaystyle \text{Putting }x=3y\ \text{or, }y=\frac{x}{3}\ \text{in }V=\frac{4}{3}\pi y^{3}+\frac{2}{3}x^{3},\ \text{we obtain}
\displaystyle V=\frac{4}{3}\pi\left(\frac{x}{3}\right)^{3}+\frac{2}{3}x^{3}=\frac{2}{3}x^{3}\left(1+\frac{2\pi}{27}\right)
\displaystyle \text{Hence, the sum of the volume is minimum when }x=3y\text{ i.e. }x\text{ is equal to three times the radius}
\displaystyle \text{of the sphere and the maximum value of the sum of the volumes is }V=\frac{2}{3}x^{3}\left(1+\frac{2\pi}{27}\right).

\displaystyle \textbf{Question 11: }\ \text{A figure consists of a semi-circle with a rectangle on its diameter. Given} \\ \text{the perimeter of}  \text{the figure, find its dimensions in order that the area may be maximum.} \\ \text{[CBSE 2002]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }ABCD\text{ be a rectangle and let the semi-circle be described on side }AB\text{ as diameter.}
\displaystyle \text{Let }AB=2x\ \text{and }AD=2y.\ \text{Let }P\text{ be the perimeter and }A\text{ be the area of the figure. Then,}
\displaystyle P=2x+4y+\pi x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{and,}\ \ A=(2x)(2y)+\frac{\pi x^{2}}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \text{Now}\ \ \ A=4xy+\frac{\pi x^{2}}{2}
\displaystyle \Rightarrow\ A=x(P-2x-\pi x)+\frac{\pi x^{2}}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ A=Px-2x^{2}-\pi x^{2}+\frac{\pi x^{2}}{2}
\displaystyle \Rightarrow\ A=Px-2x^{2}-\frac{\pi x^{2}}{2}
\displaystyle \Rightarrow\ \frac{dA}{dx}=P-4x-\pi x\ \text{and}\ \frac{d^{2}A}{dx^{2}}=-4-\pi
\displaystyle \text{The critical numbers of }A\text{ are given by }\frac{dA}{dx}=0.
\displaystyle \therefore\ \frac{dA}{dx}=0\Rightarrow P-4x-\pi x=0\Rightarrow x=\frac{P}{\pi+4}
\displaystyle \text{Clearly,}\ \frac{d^{2}A}{dx^{2}}=-4-\pi<0\ \text{for all values of }x.\ \text{Thus, }A\ \text{is maximum when }x=\frac{P}{\pi+4}.
\displaystyle \text{Putting }x=\frac{P}{\pi+4}\ \text{in (i) we get }y=\frac{P}{2(\pi+4)}.
\displaystyle \text{So, area of the figure is maximum when dimensions of the figure are:}
\displaystyle \text{Length }=2x=\frac{2P}{\pi+4}\ \ \ \ \text{and}\ \ \ \ \text{Breadth }=2y=\frac{P}{\pi+4}.

\displaystyle \textbf{Question 12: }\ \text{Find the volume of the largest cylinder that can be inscribed} \\ \text{in a sphere of radius }r\ \text{cm.}\ \text{[CBSE 2009, 2012]}
\displaystyle \text{Answer:} \displaystyle \text{Let }h\text{ be the height and }R\text{ be the radius of the base of the inscribed cylinder. Let } \\ V\text{ be the volume of the cylinder. Then,}
\displaystyle V=\pi R^{2}h\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Applying Pythagoras Theorem in }\triangle OCA,\ \text{we get}
\displaystyle OA^{2}=OC^{2}+CA^{2}
\displaystyle \Rightarrow\ r^{2}=\left(\frac{h}{2}\right)^{2}+R^{2}\Rightarrow R^{2}=r^{2}-\frac{h^{2}}{4}
\displaystyle \text{Substituting the value of }R^{2}\ \text{in (i), we get}
\displaystyle V=\pi\left(r^{2}-\frac{h^{2}}{4}\right)h
\displaystyle \Rightarrow\ V=\pi r^{2}h-\frac{\pi h^{3}}{4}
\displaystyle \Rightarrow\ \frac{dV}{dh}=\pi r^{2}-\frac{3\pi h^{2}}{4}\ \text{and}\ \frac{d^{2}V}{dh^{2}}=-\frac{3\pi h}{2}
\displaystyle \text{The critical numbers of }V\text{ are given by }\frac{dV}{dh}=0.
\displaystyle \therefore\ \frac{dV}{dh}=0\Rightarrow \pi r^{2}-\frac{3\pi h^{2}}{4}=0\Rightarrow h^{2}=\frac{4r^{2}}{3}\Rightarrow h=\frac{2r}{\sqrt{3}}
\displaystyle \text{Clearly,}\ \left(\frac{d^{2}V}{dh^{2}}\right)_{h=\frac{2r}{\sqrt{3}}}=-\sqrt{3}\pi r<0.\ \text{Thus, }V\text{ is maximum when }h=\frac{2r}{\sqrt{3}}.
\displaystyle \text{Putting }h=\frac{2r}{\sqrt{3}}\ \text{in }R^{2}=r^{2}-\frac{h^{2}}{4},\ \text{we obtain }R=r\sqrt{\frac{2}{3}}.
\displaystyle \text{Substituting the values of }R^{2}\ \text{and }h\ \text{in (i), we find that the maximum volume of the cylinder is}
\displaystyle V=\pi R^{2}h=\pi\left(\frac{2r^{2}}{3}\right)\left(\frac{2r}{\sqrt{3}}\right)=\frac{4\pi r^{3}}{3\sqrt{3}}.

\displaystyle \textbf{Question 13: }\ \text{Show that a cylinder of a given volume which is open at the top, has} \\ \text{minimum total surface} \text{area, provided its height is equal to the radius of its base.} \\ \text{[CBSE 2011, 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }r\text{ be the radius and }h\text{ be the height of a cylinder of given volume }V.\ \text{Then,}
\displaystyle V=\pi r^{2}h\Rightarrow h=\frac{V}{\pi r^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }S\text{ be the total surface area of the cylinder which is open at the top. Then,}
\displaystyle S=2\pi rh+\pi r^{2}
\displaystyle \Rightarrow\ S=2\pi r\times\frac{V}{\pi r^{2}}+\pi r^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ S=\frac{2V}{r}+\pi r^{2}
\displaystyle \Rightarrow\ \frac{dS}{dr}=-\frac{2V}{r^{2}}+2\pi r\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \text{The critical numbers of }S\text{ are given by }\frac{dS}{dr}=0.
\displaystyle \therefore\ \frac{dS}{dr}=0\Rightarrow -\frac{2V}{r^{2}}+2\pi r=0\Rightarrow V=\pi r^{3}\Rightarrow \pi r^{2}h=\pi r^{3}\Rightarrow h=r\ \ [\because\ V=\pi r^{2}h]
\displaystyle \text{Differentiating (ii) with respect to }r,\ \text{we get}
\displaystyle \frac{d^{2}S}{dr^{2}}=\frac{4V}{r^{3}}+2\pi
\displaystyle \left(\frac{d^{2}S}{dr^{2}}\right)_{r=h}=\frac{4V}{h^{3}}+2\pi>0.
\displaystyle \text{Hence, }S\text{ is minimum when }h=r\ \text{i.e., when the height of the cylinder is equal to the} \\ \text{radius of} \text{the base.}

\displaystyle \textbf{Question 14: }\ \text{Show that the height of the closed cylinder of given surface and maximum} \\ \text{volume, is equal} \text{to the diameter of its base.}\ \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }r\text{ be the radius of the base and }h\text{ be the height of a closed cylinder of given} \\ \text{surface area }S.\ \text{Then,}
\displaystyle S=2\pi r^{2}+2\pi rh
\displaystyle \Rightarrow\ h=\frac{S-2\pi r^{2}}{2\pi r}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }V\text{ be the volume of the cylinder. Then,}
\displaystyle V=\pi r^{2}h
\displaystyle \Rightarrow\ V=\pi r^{2}\left(\frac{S-2\pi r^{2}}{2\pi r}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ V=\frac{Sr}{2}-\pi r^{3}
\displaystyle \Rightarrow\ \frac{dV}{dr}=\frac{S}{2}-3\pi r^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \text{The critical numbers of }V\text{ are given by }\frac{dV}{dr}=0.
\displaystyle \text{Now,}\ \frac{dV}{dr}=0\Rightarrow \frac{S}{2}-3\pi r^{2}=0\Rightarrow S=6\pi r^{2}
\displaystyle \Rightarrow\ 2\pi r^{2}+2\pi rh=6\pi r^{2}\Rightarrow h=2r.
\displaystyle \text{Differentiating (ii) with respect to }r,\ \text{we obtain}
\displaystyle \frac{d^{2}V}{dr^{2}}=-6\pi r<0\ \text{for all }r.
\displaystyle \text{Hence, }V\text{ is maximum when }h=2r\ \text{i.e., when the height of the cylinder is equal to the diameter}
\displaystyle \text{of the base.}

\displaystyle \textbf{Question 15: }\ \text{Show that the height of a cylinder, which is open at the top, having a} \\ \text{given surface area} \text{and greatest volume, is equal to the radius of its base.}\ \text{[CBSE 2004, 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }r\text{ be the radius and }h\text{ be the height of a cylinder of given surface }S.\ \text{Then,}
\displaystyle S=\pi r^{2}+2\pi rh
\displaystyle \Rightarrow\ h=\frac{S-\pi r^{2}}{2\pi r}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }V\text{ be the volume of the cylinder. Then,}
\displaystyle V=\pi r^{2}h
\displaystyle \Rightarrow\ V=\pi r^{2}\left(\frac{S-\pi r^{2}}{2\pi r}\right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using (i)}]
\displaystyle \Rightarrow\ V=\frac{Sr}{2}-\frac{\pi r^{3}}{2}
\displaystyle \Rightarrow\ \frac{dV}{dr}=\frac{S}{2}-\frac{3\pi r^{2}}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \text{The critical numbers of }V\text{ are given by }\frac{dV}{dr}=0.
\displaystyle \therefore\ \frac{dV}{dr}=0\Rightarrow \frac{S}{2}-\frac{3\pi r^{2}}{2}=0\Rightarrow S=3\pi r^{2}
\displaystyle \Rightarrow\ \pi r^{2}+2\pi rh=3\pi r^{2}\Rightarrow r=h.
\displaystyle \text{Differentiating (ii) with respect to }r,\ \text{we get}
\displaystyle \frac{d^{2}V}{dr^{2}}=-3\pi r<0.
\displaystyle \text{Hence, }V\text{ is maximum when }r=h\ \text{i.e., when the height of the cylinder is equal to the} \\ \text{radius} \text{of its base.}

\displaystyle \textbf{Question 16: }\ \text{Show that the height of the cylinder of maximum volume that can be} \\ \text{inscribed in a sphere} \text{of radius }a\text{ is }\frac{2a}{\sqrt{3}}.\ \text{[CBSE 2001, 2012, 2013, 2014]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }r\text{ be the radius of the base and }h\text{ be the height of the cylinder }ABCD
\displaystyle \text{which is inscribed in a sphere of radius }a.\ \text{It is obvious that for maximum volume}
\displaystyle \text{the axis of the cylinder must be along the diameter of the sphere. Let }O\text{ be the centre}
\displaystyle \text{of the sphere such that }OL=x.\ \text{By symmetry, }O\text{ is the mid-point of }LM.
\displaystyle \text{Applying Pythagoras Theorem in }\triangle ALO,\ \text{we get}
\displaystyle OA^{2}=OL^{2}+AL^{2}
\displaystyle \Rightarrow\ a^{2}=x^{2}+AL^{2}
\displaystyle \Rightarrow\ AL=\sqrt{a^{2}-x^{2}}
\displaystyle \text{Let }V\text{ be the volume of the cylinder. Then,}
\displaystyle V=\pi (AL)^{2}\times LM
\displaystyle \Rightarrow\ V=\pi (AL)^{2}\times 2(OL)
\displaystyle \Rightarrow\ V=\pi (a^{2}-x^{2})\times 2x
\displaystyle \Rightarrow\ V=2\pi (a^{2}x-x^{3})
\displaystyle \Rightarrow\ \frac{dV}{dx}=2\pi (a^{2}-3x^{2})\ \text{and}\ \frac{d^{2}V}{dx^{2}}=-12\pi x
\displaystyle \text{The critical numbers of }V\text{ are given by }\frac{dV}{dx}=0.
\displaystyle \therefore\ \frac{dV}{dx}=0\Rightarrow 2\pi (a^{2}-3x^{2})=0\Rightarrow x=\frac{a}{\sqrt{3}}
\displaystyle \text{Clearly,}\ \left(\frac{d^{2}V}{dx^{2}}\right)_{x=\frac{a}{\sqrt{3}}}=-12\pi \frac{a}{\sqrt{3}}<0.
\displaystyle \text{Hence, }V\text{ is maximum when }x=\frac{a}{\sqrt{3}}\ \text{and hence }LM=2x=\frac{2a}{\sqrt{3}}.
\displaystyle \text{In other words, the height of the cylinder of maximum volume is }\frac{2a}{\sqrt{3}}.

\displaystyle \textbf{Question 17: }\ \text{Show that the semi-vertical angle of a cone of maximum volume and} \\ \text{given slant height is} \tan^{-1}\sqrt{2}\ \text{or }\cos^{-1}\frac{1}{\sqrt{3}}.\ \text{[CBSE 2011, 2014]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }\alpha\text{ be the semi-vertical angle of a cone }VAB\text{ of given slant height }l.
\displaystyle \text{In }\triangle AOV,\ \cos\alpha=\frac{VO}{VA}\ \text{and}\ \sin\alpha=\frac{OA}{VA}
\displaystyle \Rightarrow\ \cos\alpha=\frac{VO}{l}\ \text{and}\ \sin\alpha=\frac{OA}{l}
\displaystyle \Rightarrow\ VO=l\cos\alpha,\ OA=l\sin\alpha
\displaystyle \text{Let }V\text{ be the volume of the cone. Then,}
\displaystyle V=\frac{1}{3}\pi (OA)^{2}(VO)
\displaystyle \Rightarrow\ V=\frac{1}{3}\pi (l\sin\alpha)^{2}(l\cos\alpha)
\displaystyle \Rightarrow\ V=\frac{1}{3}\pi l^{3}\sin^{2}\alpha\cos\alpha
\displaystyle \Rightarrow\ \frac{dV}{d\alpha}=\frac{\pi l^{3}}{3}\left(-\sin^{3}\alpha+2\sin\alpha\cos^{2}\alpha\right)
\displaystyle \Rightarrow\ \frac{dV}{d\alpha}=\frac{\pi l^{3}}{3}\sin\alpha\left(-\sin^{2}\alpha+2\cos^{2}\alpha\right)\ \ \ \ \ ...(i)
\displaystyle \text{The critical points of }V\text{ are given by }\frac{dV}{d\alpha}=0.
\displaystyle \therefore\ \frac{dV}{d\alpha}=0\Rightarrow \frac{\pi l^{3}}{3}\sin\alpha\left(-\sin^{2}\alpha+2\cos^{2}\alpha\right)=0
\displaystyle \Rightarrow\ 2\cos^{2}\alpha=\sin^{2}\alpha
\displaystyle \Rightarrow\ \tan^{2}\alpha=2\Rightarrow \tan\alpha=\sqrt{2}\ \ [\because\ \alpha\text{ is acute : }\sin\alpha\neq 0]
\displaystyle \Rightarrow\ \cos\alpha=\frac{1}{\sqrt{1+\tan^{2}\alpha}}=\frac{1}{\sqrt{3}}\ \ \ \ \ [\because\ \tan\alpha=\sqrt{2}]
\displaystyle \text{Differentiating (i) with respect to }\alpha,\ \text{we get}
\displaystyle \frac{d^{2}V}{d\alpha^{2}}=\frac{\pi l^{3}}{3}\left(2\cos^{3}\alpha(2-7\tan^{2}\alpha)\right)
\displaystyle \left(\frac{d^{2}V}{d\alpha^{2}}\right)_{\tan\alpha=\sqrt{2}}=\frac{\pi l^{3}}{3}\left(\frac{1}{\sqrt{3}}\right)^{3}(2-7\times 2)=-\frac{4\pi l^{3}}{3\sqrt{3}}<0.
\displaystyle \text{Thus, }V\text{ is maximum, when }\tan\alpha=\sqrt{2},\ \text{or, }\alpha=\tan^{-1}\sqrt{2}\ \text{i.e., when the semi-vertical angle of the cone is }\tan^{-1}\sqrt{2}.

\displaystyle \textbf{Question 18: }\ \text{Show that the volume of the largest cone that can be inscribed in a} \\ \text{sphere of radius }R \text{is }\frac{8}{27}\ \text{of the volume of the sphere.} \\ \text{[CBSE 2008, 2010 C, 2012, 2013, 2014, 2016]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }VAB\text{ be a cone of greatest volume inscribed in a sphere of radius }R.
\displaystyle \text{It is obvious that for maximum volume the axis of the cone must be along the diameter of the sphere.}
\displaystyle \text{Let }VC\text{ be the axis of the cone and }O\text{ be the centre of the sphere such that }OC=x.
\displaystyle VC=VO+OC=R+x=\text{height of the cone.}
\displaystyle \text{Applying Pythagoras theorem in }\triangle ACO,\ \text{we get}
\displaystyle OA^{2}=AC^{2}+OC^{2}
\displaystyle \Rightarrow\ AC^{2}=OA^{2}-OC^{2}=R^{2}-x^{2}
\displaystyle \text{Let }V\text{ be the volume of the cone. Then,}
\displaystyle V=\frac{1}{3}\pi (AC)^{2}(VC)
\displaystyle \Rightarrow\ V=\frac{1}{3}\pi (R^{2}-x^{2})(R+x)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \Rightarrow\ \frac{dV}{dx}=\frac{1}{3}\pi\left[R^{2}-x^{2}-2x(R+x)\right]
\displaystyle \Rightarrow\ \frac{dV}{dx}=\frac{1}{3}\pi (R^{2}-2Rx-3x^{2})\ \text{and}\ \frac{d^{2}V}{dx^{2}}=\frac{1}{3}\pi (-2R-6x)
\displaystyle \text{The critical numbers of }V\text{ are given by }\frac{dV}{dx}=0.
\displaystyle \therefore\ \frac{dV}{dx}=0\Rightarrow R^{2}-2Rx-3x^{2}=0\Rightarrow (R-3x)(R+x)=0
\displaystyle \Rightarrow\ R-3x=0\Rightarrow x=\frac{R}{3}\ \ \ \ \ [\because\ R+x\neq 0]
\displaystyle \text{Putting }x=\frac{R}{3}\ \text{in }\frac{d^{2}V}{dx^{2}}=\frac{1}{3}\pi (-2R-6x),\ \text{we get}
\displaystyle \left(\frac{d^{2}V}{dx^{2}}\right)_{x=\frac{R}{3}}=-\frac{4}{3}\pi R<0.
\displaystyle \text{Thus, }V\text{ is maximum when }x=\frac{R}{3}.
\displaystyle \text{Putting }x=\frac{R}{3}\ \text{in (i), we obtain}
\displaystyle V=\frac{1}{3}\pi\left(R^{2}-\frac{R^{2}}{9}\right)\left(R+\frac{R}{3}\right)=\frac{32\pi R^{3}}{81}
\displaystyle =\frac{8}{27}\left(\frac{4}{3}\pi R^{3}\right)=\frac{8}{27}\ \text{(Volume of the sphere).}

\displaystyle \textbf{Question 19: }\ \text{Prove that the radius of the right circular cylinder of greatest} \\ \text{curved surface which can be} \text{inscribed in a given cone is half of that of the cone.} \\ \text{[CBSE 2010 C, 2012, 2013]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }VAB\text{ be the cone of base radius }r=OA\ \text{and height }h=VO.
\displaystyle \text{Let a cylinder of base radius }OC=x\ \text{and height }OO'\ \text{be inscribed in the cone.}
\displaystyle \text{Clearly, }\triangle VOB\sim \triangle A'B'DB
\displaystyle \therefore\ \frac{VO}{B'D}=\frac{OB}{DB}
\displaystyle \Rightarrow\ \frac{h}{B'D}=\frac{r}{r-x}
\displaystyle \Rightarrow\ B'D=\frac{h(r-x)}{r}
\displaystyle \text{Let }S\text{ be the curved surface area of the cylinder. Then,}
\displaystyle S=2\pi(OC)(B'D)
\displaystyle \Rightarrow\ S=2\pi x\frac{h(r-x)}{r}=\frac{2\pi h}{r}(rx-x^{2})
\displaystyle \Rightarrow\ \frac{dS}{dx}=\frac{2\pi h}{r}(r-2x)\ \text{and}\ \frac{d^{2}S}{dx^{2}}=-\frac{4\pi h}{r}
\displaystyle \text{The critical numbers of }S\ \text{are given by }\frac{dS}{dx}=0.
\displaystyle \therefore\ \frac{dS}{dx}=0\Rightarrow \frac{2\pi h}{r}(r-2x)=0\Rightarrow x=\frac{r}{2}
\displaystyle \text{Clearly,}\ \frac{d^{2}S}{dx^{2}}=-\frac{4\pi h}{r}<0\ \text{for all }x.
\displaystyle \text{Hence, }S\text{ is maximum when }x=\frac{r}{2}\ \text{i.e. radius of the cylinder is half of the radius of the cone.}

\displaystyle \textbf{Question 20: }\ \text{Show that the volume of the greatest cylinder which can be inscribed} \\ \text{in a cone of height }h  \text{and semi-vertical angle }\alpha\ \text{is }\frac{4}{27}\pi h^{3}\tan^{2}\alpha.\ \text{Also, show that} \\ \text{height of the cylinder is }\frac{h}{3}.  \text{[CBSE 2001, 2007, 2008, 2010, 2017]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }VAB\text{ be a given cone of height }h,\ \text{semi-vertical angle }\alpha
\displaystyle \text{and let }x\text{ be the radius of the base of the cylinder }A'B'DC\ \text{which is inscribed in the cone }VAB.
\displaystyle \text{In }\triangle VO'A',\ \tan\alpha=\frac{O'A'}{VO'}=\frac{x}{VO'}
\displaystyle \Rightarrow\ VO'=x\cot\alpha
\displaystyle \Rightarrow\ OO'=VO-VO'=h-x\cot\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{Let }V\text{ be the volume of the cylinder. Then,}
\displaystyle V=\pi (O'B')^{2}(OO')
\displaystyle \Rightarrow\ V=\pi x^{2}(h-x\cot\alpha)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \Rightarrow\ \frac{dV}{dx}=2\pi xh-3\pi x^{2}\cot\alpha
\displaystyle \text{The critical numbers of }V\text{ are given by }\frac{dV}{dx}=0.
\displaystyle \therefore\ \frac{dV}{dx}=0\Rightarrow 2\pi xh-3\pi x^{2}\cot\alpha=0
\displaystyle \Rightarrow\ x=\frac{2h}{3}\tan\alpha\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\because\ x\neq 0]
\displaystyle \text{Now,}\ \frac{d^{2}V}{dx^{2}}=2\pi h-6\pi x\cot\alpha
\displaystyle \text{When }x=\frac{2h}{3}\tan\alpha,
\displaystyle \frac{d^{2}V}{dx^{2}}=\pi(2h-4h)=-2\pi h<0.
\displaystyle \text{Hence, }V\text{ is maximum when }x=\frac{2h}{3}\tan\alpha.
\displaystyle \text{Putting }x=\frac{2h}{3}\tan\alpha\ \text{in (ii), the maximum volume of the cylinder is given by}
\displaystyle V=\pi\left(\frac{2h}{3}\tan\alpha\right)^{2}\left(h-\frac{2h}{3}\right)=\frac{4}{27}\pi h^{3}\tan^{2}\alpha.
\displaystyle \text{Putting }x=\frac{2h}{3}\tan\alpha\ \text{in (i), we get}
\displaystyle OO'=h-x\cot\alpha=h-\frac{2h}{3}=\frac{h}{3}.
\displaystyle \text{Hence, height of the cylinder }=OO'=\frac{h}{3}.

\displaystyle \textbf{Question 21: }\ \text{Let }AP\ \text{and }BQ\ \text{be two vertical poles at points }A\ \text{and }B \\ \text{respectively. If }AP=16\text{ m,}  BQ=22\text{ m and }AB=20\text{ m, then find the distance} \\ \text{of a point }R\ \text{on }AB\ \text{from the point }A\ \text{such that} RP^{2}+RQ^{2}\ \text{is minimum.}\ \text{[CBSE 2010]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }R\text{ be a point on }AB\ \text{such that }AR=x\text{ m. Then, }RB=(20-x)\text{ m.}
\displaystyle \text{Applying Pythagoras Theorem in }\triangle RAP\ \text{and }\triangle RBQ,\ \text{we get}
\displaystyle PR^{2}=x^{2}+16^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \text{and,}\ \ RQ^{2}=22^{2}+(20-x)^{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(ii)
\displaystyle \therefore\ PR^{2}+RQ^{2}=x^{2}+16^{2}+22^{2}+(20-x)^{2}=2x^{2}-40x+1140
\displaystyle \text{Let }Z=RP^{2}+RQ^{2}.\ \text{Then,}
\displaystyle Z=2x^{2}-40x+1140
\displaystyle \Rightarrow\ \frac{dZ}{dx}=4x-40\ \text{and}\ \frac{d^{2}Z}{dx^{2}}=4
\displaystyle \text{The critical numbers of }Z\ \text{are given by }\frac{dZ}{dx}=0.
\displaystyle \therefore\ \frac{dZ}{dx}=0\Rightarrow 4x-40=0\Rightarrow x=10
\displaystyle \text{Clearly,}\ \frac{d^{2}Z}{dx^{2}}=4>0\ \text{for all }x.\ \text{So, }Z\ \text{is minimum when }x=10.
\displaystyle \text{Thus, }RP^{2}+RQ^{2}\ \text{is minimum when, the distance of }R\ \text{from }A\ \text{is }10\text{ m.}

\displaystyle \textbf{Question 22: }\ \text{If the length of three sides of a trapezium other than base are equal to } \\ 10\text{ cm,}  \text{then find the area of trapezium when it is maximum.}\ \text{[CBSE 2010, 2013]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }ABCD\text{ be the given trapezium such that }AD=DC=BC=10\text{ cm. Draw }DP
\displaystyle \text{and }CQ\ \text{perpendiculars from }D\ \text{and }C\ \text{respectively on }AB.
\displaystyle \text{Clearly, }\triangle APD\cong \triangle BQC.
\displaystyle \text{Let }AP=x\text{ cm. Then, }BQ=x\text{ cm.}
\displaystyle \text{By applying Pythagoras Theorem in }\triangle APD\ \text{and }\triangle BQC,\ \text{we obtain}
\displaystyle DP=QC=\sqrt{100-x^{2}}
\displaystyle \text{Let }A\text{ be the area of trapezium }ABCD.\ \text{Then,}
\displaystyle A=\frac{1}{2}(AB+CD)\times DP
\displaystyle \Rightarrow\ A=\frac{1}{2}(10+10+2x)\sqrt{100-x^{2}}
\displaystyle \Rightarrow\ A=(10+x)\sqrt{100-x^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \Rightarrow\ \frac{dA}{dx}=\sqrt{100-x^{2}}-\frac{x(10+x)}{\sqrt{100-x^{2}}}=\frac{100-10x-2x^{2}}{\sqrt{100-x^{2}}}
\displaystyle \text{The critical numbers of }A\text{ are given by }\frac{dA}{dx}=0.
\displaystyle \therefore\ \frac{dA}{dx}=0\Rightarrow \frac{100-10x-2x^{2}}{\sqrt{100-x^{2}}}=0
\displaystyle \Rightarrow\ 100-10x-2x^{2}=0
\displaystyle \Rightarrow\ x^{2}+5x-50=0
\displaystyle \Rightarrow\ (x+10)(x-5)=0
\displaystyle \Rightarrow\ x=5\ \ \ \ \ [\because\ x>0\Rightarrow x+10\neq 0]
\displaystyle \text{Now,}\ \frac{d^{2}A}{dx^{2}}=\frac{2x^{3}-300x-1000}{(100-x^{2})^{3/2}}
\displaystyle \left(\frac{d^{2}A}{dx^{2}}\right)_{x=5}=-\frac{30}{\sqrt{75}}<0.
\displaystyle \text{Thus, the area of the trapezium is maximum when }x=5.
\displaystyle \text{Putting }x=5\ \text{in (i), the maximum area is given by}
\displaystyle A=\frac{1}{2}(10+5)\sqrt{100-25}=\frac{75\sqrt{3}}{2}\ \text{cm}^{2}.

\displaystyle \textbf{Question 23: }\ \text{Find the area of the greatest isosceles triangle that can be inscribed in a} \\ \text{given ellipse} \text{having its vertex coincident with one end of the major axis.}\ \text{[CBSE 2010]}
\displaystyle \text{Answer:} \displaystyle  \text{Let the equation of the ellipse be }\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.
\displaystyle \text{Let }APQ\ \text{be an isosceles triangle having one vertex at }A(a,0).\ \text{Let the coordinates of }P\ \text{be}
\displaystyle (a\cos\theta,\ b\sin\theta).\ \text{Then the coordinates of }Q\ \text{are }(a\cos\theta,\ -b\sin\theta).
\displaystyle \text{Let }A\ \text{be the area of }\triangle APQ.\ \text{Then,}
\displaystyle A=\frac{1}{2}(PQ)(AM)
\displaystyle \Rightarrow\ A=\frac{1}{2}(2b\sin\theta)(a-a\cos\theta)
\displaystyle \Rightarrow\ A=ab(\sin\theta-\sin\theta\cos\theta)
\displaystyle \Rightarrow\ \frac{dA}{d\theta}=ab(\cos\theta-\cos^{2}\theta+\sin^{2}\theta)
\displaystyle \Rightarrow\ \frac{dA}{d\theta}=ab(\cos\theta-\cos 2\theta)
\displaystyle \text{The critical numbers of }A\ \text{are given by }\frac{dA}{d\theta}=0.
\displaystyle \therefore\ \frac{dA}{d\theta}=0
\displaystyle \Rightarrow\ ab(\cos\theta-\cos 2\theta)=0
\displaystyle \Rightarrow\ \cos\theta=\cos 2\theta
\displaystyle \Rightarrow\ \theta=2\pi-2\theta
\displaystyle \Rightarrow\ \theta=\frac{2\pi}{3}
\displaystyle \text{Now,}\ \frac{dA}{d\theta}=ab(\cos\theta-\cos 2\theta)
\displaystyle \Rightarrow\ \frac{d^{2}A}{d\theta^{2}}=ab(-\sin\theta+2\sin 2\theta)
\displaystyle \text{For }\theta=\frac{2\pi}{3},\ \text{we obtain}
\displaystyle \frac{d^{2}A}{d\theta^{2}}=ab\left(-\sin\frac{2\pi}{3}+2\sin\frac{4\pi}{3}\right)=ab\left(-\frac{\sqrt{3}}{2}-2\times\frac{\sqrt{3}}{2}\right)<0
\displaystyle \text{Hence, }A\ \text{is maximum when }\theta=\frac{2\pi}{3}.\ \text{The maximum area }A\ \text{is given by}
\displaystyle A=ab\left(\sin\frac{2\pi}{3}-\sin\frac{2\pi}{3}\cos\frac{2\pi}{3}\right)=ab\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times\frac{1}{2}\right)=\frac{3\sqrt{3}}{4}\ ab.

\displaystyle \textbf{Question 24: }\ \text{Find the area of the greatest rectangle that can be inscribed in an} \\ \text{ellipse } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1. \text{[CBSE 2013]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }PQRS\text{ be a rectangle inscribed in the ellipse }\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1.
\displaystyle \text{Let the coordinate of }P\text{ be }(a\cos\theta,\ b\sin\theta).
\displaystyle \text{Then, the coordinates of }Q,\ R\text{ and }S\text{ are }(-a\cos\theta,\ b\sin\theta),
\displaystyle (-a\cos\theta,\ -b\sin\theta)\ \text{and }(a\cos\theta,\ -b\sin\theta)\ \text{respectively.}
\displaystyle \text{Let }A\text{ be the area of rectangle }PQRS.\ \text{Then,}
\displaystyle A=PQ\times PS
\displaystyle \Rightarrow\ A=2a\cos\theta\times 2b\sin\theta
\displaystyle \Rightarrow\ A=2ab\sin 2\theta\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...(i)
\displaystyle \Rightarrow\ \frac{dA}{d\theta}=4ab\cos 2\theta\ \text{and}\ \frac{d^{2}A}{d\theta^{2}}=-8ab\sin 2\theta
\displaystyle \text{The critical numbers of }A\text{ are given by }\frac{dA}{d\theta}=0.
\displaystyle \therefore\ \frac{dA}{d\theta}=0\Rightarrow 4ab\cos 2\theta=0\Rightarrow \cos 2\theta=0
\displaystyle \Rightarrow\ 2\theta=\frac{\pi}{2}\ \text{or}\ \frac{3\pi}{2}\Rightarrow \theta=\frac{\pi}{4}\ \text{or}\ \theta=\frac{3\pi}{4}
\displaystyle \text{Clearly, }\left(\frac{d^{2}A}{d\theta^{2}}\right)_{\theta=\frac{\pi}{4}}=-8ab\sin\frac{\pi}{2}=-8ab<0.
\displaystyle \text{So, }A\text{ is maximum when }\theta=\frac{\pi}{4}.
\displaystyle \text{Putting }\theta=\frac{\pi}{4}\ \text{in (i), the maximum value of }A\text{ is given by}
\displaystyle A=2ab\sin\frac{\pi}{2}=2ab.
\displaystyle \text{Hence, the area of the greatest rectangle is }2ab\ \text{sq. units.}

\displaystyle \textbf{Question 25: }\ \text{A point on the hypotenuse of a right triangle is at distances }a\text{ and }b \\ \text{from the sides of the}   \text{triangle. Show that the minimum length of the hypotenuse is } \\ \left(a^{2/3}+b^{2/3}\right)^{3/2}.  \text{[CBSE 2008]}
\displaystyle \text{Answer:} \displaystyle  \text{Let }AOB\text{ be a right triangle with hypotenuse }AB\ \text{such that a point }P\ \text{on }AB\ \text{is at}
\displaystyle \text{distances }a\ \text{and }b\ \text{from }OA\ \text{and }OB\ \text{respectively i.e. }PL=a\ \text{and }PM=b.
\displaystyle \text{Let }\angle OAB=\theta.\ \text{In }\triangle ALP\ \text{and }\triangle PMB
\displaystyle \sin\theta=\frac{PL}{AP}\ \text{and}\ \cos\theta=\frac{PM}{BP}
\displaystyle \Rightarrow\ \sin\theta=\frac{a}{AP}\ \text{and}\ \cos\theta=\frac{b}{BP}
\displaystyle \Rightarrow\ AP=a\,\mathrm{cosec}\,\theta\ \text{and}\ BP=b\sec\theta
\displaystyle \text{Let }l\ \text{be the length of the hypotenuse }AB.\ \text{Then,}
\displaystyle l=AP+BP
\displaystyle \Rightarrow\ l=a\,\mathrm{cosec}\,\theta+b\sec\theta
\displaystyle \Rightarrow\ \frac{dl}{d\theta}=-a\,\mathrm{cosec}\,\theta\cot\theta+b\sec\theta\tan\theta
\displaystyle \text{and,}\ \frac{d^{2}l}{d\theta^{2}}=a\,\mathrm{cosec}^{3}\theta+a\,\mathrm{cosec}\,\theta\cot^{2}\theta+b\sec^{3}\theta+b\sec\theta\tan^{2}\theta
\displaystyle \text{The critical numbers of }l\ \text{are given by }\frac{dl}{d\theta}=0.
\displaystyle \therefore\ \frac{dl}{d\theta}=0
\displaystyle \Rightarrow\ -a\,\mathrm{cosec}\,\theta\cot\theta+b\sec\theta\tan\theta=0
\displaystyle \Rightarrow\ -\frac{a\cos\theta}{\sin^{2}\theta}+\frac{b\sin\theta}{\cos^{2}\theta}=0
\displaystyle \Rightarrow\ \tan^{3}\theta=\frac{a}{b}
\displaystyle \Rightarrow\ \tan\theta=\left(\frac{a}{b}\right)^{1/3}
\displaystyle \Rightarrow\ \sin\theta=\frac{a^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}\ \text{and,}\ \cos\theta=\frac{b^{1/3}}{\sqrt{a^{2/3}+b^{2/3}}}
\displaystyle \text{Clearly,}\ \frac{d^{2}l}{d\theta^{2}}>0\ \text{for }\tan\theta=\left(\frac{a}{b}\right)^{1/3}.\ \text{Thus, }l\ \text{is minimum when }\tan\theta=\left(\frac{a}{b}\right)^{1/3}.
\displaystyle \text{The minimum value of }l\ \text{is given by}
\displaystyle l=a\,\mathrm{cosec}\,\theta+b\sec\theta=a\sqrt{1+\cot^{2}\theta}+b\sqrt{1+\tan^{2}\theta}=a\sqrt{1+\left(\frac{b}{a}\right)^{2/3}}+b\sqrt{1+\left(\frac{a}{b}\right)^{2/3}}
\displaystyle \Rightarrow\ l=\left(a^{2/3}+b^{2/3}\right)^{3/2}.

\displaystyle \textbf{Question 26. }\text{Find the maximum and minimum values of the function given by } \\ f(x)=5+\sin 2x. \hspace{3.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=5+\sin 2x
\displaystyle \Rightarrow f'(x)=2\cos 2x
\displaystyle \Rightarrow f''(x)=-4\sin 2x
\displaystyle \text{For maxima or minima, put }f'(x)=0
\displaystyle \Rightarrow 2\cos 2x=0
\displaystyle \Rightarrow \cos 2x=0=\cos\frac{\pi}{2}\text{ or }\cos\frac{3\pi}{2}
\displaystyle \Rightarrow 2x=\frac{\pi}{2},\frac{3\pi}{2}\Rightarrow x=\frac{\pi}{4},\frac{3\pi}{4}
\displaystyle \text{Now, }f''\left(\frac{\pi}{4}\right)=-4\sin\left(2\times\frac{\pi}{4}\right)
\displaystyle =-4\sin\frac{\pi}{2}=-4<0
\displaystyle \text{So, }x=\frac{\pi}{4}\text{ is point of maxima.}
\displaystyle \text{and }f''\left(\frac{3\pi}{4}\right)=-4\sin\left(2\times\frac{3\pi}{4}\right)=-4\sin\frac{3\pi}{2}
\displaystyle =-4\sin\left(2\pi-\frac{\pi}{2}\right)=4\sin\frac{\pi}{2}=4>0
\displaystyle \therefore x=\frac{3\pi}{4}\text{ is the point of minima.}
\displaystyle \therefore \text{Maximum value of }f(x)\text{ at }x=\frac{\pi}{4}\text{ is }5+\sin\left(2\times\frac{\pi}{4}\right)
\displaystyle =5+\sin\left(\frac{\pi}{2}\right)=5+1=6
\displaystyle \text{and minimum value of }f(x)\text{ at }x=\frac{3\pi}{4}
\displaystyle \text{is }5+\sin\left(2\times\frac{3\pi}{4}\right)=5+\sin\frac{3\pi}{2}=5-\sin\frac{\pi}{2}=5-1=4
\\

\displaystyle \textbf{Question 27. }\text{Sum of two numbers is 5. If the sum of the cubes of these numbers} \\ \text{is least, then find the sum of the squares of these numbers.} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let one number be }x\text{, then the other number will be }(5-x)
\displaystyle \text{Let the sum of the cubes of these number be }S
\displaystyle \therefore S=x^3+(5-x)^3
\displaystyle \Rightarrow \frac{dS}{dx}=3x^2+3(5-x)^2(-1)
\displaystyle \Rightarrow \frac{dS}{dx}=3x^2-3(5-x)^2
\displaystyle \Rightarrow \frac{d^2S}{dx^2}=6x+6(5-x)\Rightarrow \frac{d^2S}{dx^2}=30
\displaystyle \text{For minimum, put }\frac{dS}{dx}=0
\displaystyle \Rightarrow 3x^2-3(5-x)^2=0
\displaystyle \Rightarrow x^2-(25+x^2-10x)=0
\displaystyle \Rightarrow x^2-25-x^2+10x=0
\displaystyle \Rightarrow 10x=25
\displaystyle \Rightarrow x=\frac{5}{2}
\displaystyle \text{At }x=\frac{5}{2},\ \frac{d^2S}{dx^2}=30>0
\displaystyle \therefore \text{At }x=\frac{5}{2},\ S\text{ is the minimum.}
\displaystyle \therefore \text{Numbers are }\frac{5}{2},\ 5-\frac{5}{2}\text{ i.e. }\frac{5}{2},\frac{5}{2}
\displaystyle \therefore \text{The sum of squares of these numbers is }\left(\frac{5}{2}\right)^2+\left(\frac{5}{2}\right)^2
\displaystyle \text{i.e. }\frac{25}{4}+\frac{25}{4}=\frac{50}{4}=\frac{25}{2}
\\

\displaystyle \textbf{Question 28. }\text{A function }f:R\rightarrow R\text{ is defined as }f(x)=x^3+1. \\ \text{ Then, the function has} \hspace{3.2cm} \text{[CBSE 2022 (Term I)]}
\displaystyle \text{(a) no minimum value}
\displaystyle \text{(b) no maximum value}
\displaystyle \text{(c) both maximum and minimum values}
\displaystyle \text{(d) neither maximum nor minimum value}
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, }f(x)=x^3+1
\displaystyle f'(x)=3x^2
\displaystyle \text{Putting }f'(x)=0,\ 3x^2=0
\displaystyle \therefore x=0
\displaystyle f''(x)=6x=0
\displaystyle \text{As }f''(x)\text{ is equal to 0, therefore the function is neither maximum nor minimum value.}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 29. }\text{The area of a trapezium is defined by function }f\text{ and given by } \\ f(x)=(10+x)\sqrt{100-x^2},\text{ then the area when it is maximised is} \hspace{0.2cm} \text{[CBSE 2022 (Term I)]}
\displaystyle \text{(a) }75\text{ cm}^2 \qquad \text{(b) }7\sqrt{3}\text{ cm}^2 \qquad \text{(c) }75\sqrt{3}\text{ cm}^2 \qquad \text{(d) }5\text{ cm}^2
\displaystyle \text{Answer:}
\displaystyle \text{(c) Given, }f(x)=(10+x)\sqrt{100-x^2}
\displaystyle \text{Since, }A\text{ has a square root.}
\displaystyle \text{Let }Z=[f(x)]^2=(x+10)^2(100-x^2)\text{, where }f'(x)=0\text{, there }z'(x)=0
\displaystyle Z=(x+10)^2(100-x^2)
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dZ}{dx}=\frac{d}{dx}[(x+10)^2(100-x^2)]
\displaystyle \Rightarrow Z'=[2(x+10)(100-x^2)-2x(x+10)^2]
\displaystyle \Rightarrow Z'=2(x+10)[100-x^2-x(x+10)]
\displaystyle \Rightarrow Z'=2(x+10)[100-x^2-x^2-10x]
\displaystyle \Rightarrow Z'=-4(x+10)(x^2+5x-50)
\displaystyle \text{Putting }\frac{dZ}{dx}=0
\displaystyle -4(x+10)[x^2+5x+50]=0
\displaystyle \Rightarrow x=-10\text{ and }x=5
\displaystyle \text{Since, }x\text{ is length, it cannot be negative.}
\displaystyle \therefore x=5
\displaystyle \text{Now, }A=(x+10)\sqrt{100-x^2}
\displaystyle =(5+10)\sqrt{100-(5)^2}
\displaystyle =15\times5\sqrt{3}=75\sqrt{3}\text{ cm}^2
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 30. }\text{Find the least value of the function }f(x)=ax+\dfrac{b}{x} \\  (a>0,\ b>0,\ x>0). \hspace{3.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }f(x)=ax+\frac{b}{x}\ (a>0,\ b>0,\ x>0)
\displaystyle \therefore f'(x)=a-\frac{b}{x^2}\text{ and }f''(x)=\frac{2b}{x^3}
\displaystyle \text{For maxima and minima of }f(x),
\displaystyle f'(x)=0
\displaystyle \Rightarrow a-\frac{b}{x^2}=0\Rightarrow x^2=\frac{b}{a}
\displaystyle \Rightarrow x=\sqrt{\frac{b}{a}}\qquad[\because x>0\text{ (given)}]
\displaystyle \text{Again, }f''\left(\sqrt{\frac{b}{a}}\right)=\frac{2b}{\left(\frac{b}{a}\right)^{3/2}}=\frac{2a^{3/2}}{\sqrt{b}}>0
\displaystyle [\because a>0,\ b>0]
\displaystyle \text{So, }f(x)\text{ has least value at }x=\sqrt{\frac{b}{a}}
\displaystyle \therefore f_{\min}(x)=f\left(\sqrt{\frac{b}{a}}\right)=a\sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}
\displaystyle =\sqrt{ab}+\sqrt{ab}=2\sqrt{ab}
\\

\displaystyle \textbf{Question 31. }\text{Find the minimum value of }(ax+by)\text{, where }xy=c^2. \hspace{0.2cm} \text{[CBSE 2020; CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=ax+by\text{, whose minimum value is required.}
\displaystyle \text{Then, }f(x)=ax+\frac{bc^2}{x}\qquad\left[\because xy=c^2\Rightarrow y=\frac{c^2}{x}\right]
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle f'(x)=a-\frac{bc^2}{x^2}
\displaystyle \text{For maximum or minimum value of }f(x)\text{, put }f'(x)=0
\displaystyle \Rightarrow a-\frac{bc^2}{x^2}=0\Rightarrow a=\frac{bc^2}{x^2}
\displaystyle \Rightarrow x^2=\frac{bc^2}{a}\Rightarrow x=\pm\sqrt{\frac{b}{a}}\cdot c
\displaystyle \text{Now, }f''(x)=0+\frac{2bc^2}{x^3}
\displaystyle \text{At }x=+\sqrt{\frac{b}{a}}\cdot c,\ f''(x)=\frac{2bc^2}{\left(\sqrt{\frac{b}{a}}\cdot c\right)^3}=+\text{ve}
\displaystyle \text{Hence, }f(x)\text{ has minimum value at }x=\sqrt{\frac{b}{a}}\cdot c
\displaystyle \text{At }x=-\sqrt{\frac{b}{a}}\cdot c,\ f''(x)=\frac{2bc^2}{\left(-\sqrt{\frac{b}{a}}\cdot c\right)^3}=-\text{ve}
\displaystyle \text{Hence, }f(x)\text{ has maximum value at }x=-\sqrt{\frac{b}{a}}\cdot c
\displaystyle \text{When }x=\sqrt{\frac{b}{a}}\cdot c\text{, then }y=\frac{c^2}{x}=\frac{c^2}{\sqrt{\frac{b}{a}}\cdot c}=\sqrt{\frac{a}{b}}\cdot c
\displaystyle \therefore \text{Minimum value of }f(x)=a\sqrt{\frac{b}{a}}\cdot c+b\sqrt{\frac{a}{b}}\cdot c
\displaystyle =\sqrt{ab}\cdot c+\sqrt{ab}\cdot c
\displaystyle =2\sqrt{ab}\cdot c
\\

\displaystyle \textbf{Question 32. }\text{Find the dimensions of the rectangle of perimeter 36 cm which will} \\ \text{sweep out a volume as large as possible, when revolved about one of its side. Also, find the} \\ \text{maximum volume.} \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }ABCD\text{ is a rectangle with length }AD=y\text{ cm and breadth }=x\text{ cm}

\displaystyle \text{The rectangle is rotated about }AD.\text{ Let }V\text{ be the volume of the cylinder so formed}
\displaystyle \therefore V=\pi x^2y\qquad\ldots\text{(i)}
\displaystyle \text{Perimeter of rectangle}=2(x+y)
\displaystyle \Rightarrow 36=2(x+y)\Rightarrow y=18-x
\displaystyle \text{Now, }V=\pi x^2(18-x)
\displaystyle V=\pi(18x^2-x^3)
\displaystyle \Rightarrow \frac{dV}{dx}=\pi(36x-3x^2)
\displaystyle \text{For maxima or minima, put }\frac{dV}{dx}=0
\displaystyle \pi(36x-3x^2)=0\Rightarrow x=12,\ x\neq0
\displaystyle \text{Now, }\frac{d^2V}{dx^2}=\pi(36-6x)
\displaystyle \left(\frac{d^2V}{dx^2}\right)_{x=12}=\pi(36-72)=-36\pi<0
\displaystyle \therefore \text{Volume is maximum when }x=12\text{ cm}
\displaystyle y=18-x=18-12=6\text{ cm}
\displaystyle \text{Hence, the dimension of rectangle, which have maximum volume, when revolved about of its side is }12\times6
\\

\displaystyle \textbf{Question 33. }\text{A tank with rectangular base and rectangular sides, open at the top is} \\ \text{to be constructed, so that its depth is 2 m and volume is 8 m}^3\text{. If building of tank cost} \\ \text{Rs. 70 per sq mfor the base and Rs. 45 per sq m for sides. What is the cost of least} \\ \text{expensive tank?} \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ m be the length, }y\text{ m be the breadth and }h=2\text{ m be the depth of} \\ \text{the tank. Let Rs. }H\text{ be the total cost for building the tank.}
\displaystyle \text{Now, given that }h=2\text{ m}
\displaystyle \text{and volume of tank}=8\text{ m}^3
\displaystyle \text{Clearly, area of the rectangular base of the tank}
\displaystyle =\text{length}\times\text{breadth}=xy\text{ m}^2
\displaystyle \text{and the area of the four rectangular sides}
\displaystyle =2(\text{length}+\text{breadth})\times\text{height}
\displaystyle =2(x+y)\times2=4(x+y)\text{ m}^2
\displaystyle \therefore \text{Total cost, }H=70\times xy+45\times4(x+y)
\displaystyle \Rightarrow H=70xy+180(x+y)\qquad\ldots\text{(i)}
\displaystyle \text{Also, volume of tank}=8\text{ m}^3
\displaystyle \Rightarrow l\times b\times h=8\Rightarrow x\times y\times2=8\Rightarrow y=\frac{4}{x}\qquad\ldots\text{(ii)}
\displaystyle \text{On putting the value of }y\text{ from Eq. (ii) in Eq. (i), we get}
\displaystyle H=70x\times\frac{4}{x}+180\left(x+\frac{4}{x}\right)
\displaystyle \Rightarrow H=280+180\left(x+\frac{4}{x}\right)\qquad\ldots\text{(iii)}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dH}{dx}=180\left(1-\frac{4}{x^2}\right)
\displaystyle \text{For maxima or minima, put }\frac{dH}{dx}=0
\displaystyle \Rightarrow 180\left(1-\frac{4}{x^2}\right)=0\Rightarrow 1-\frac{4}{x^2}=0
\displaystyle \Rightarrow \frac{4}{x^2}=1
\displaystyle \Rightarrow x^2=4\Rightarrow x=2\qquad[\because x>0]
\displaystyle \text{Also, }\frac{d^2H}{dx^2}=\frac{d}{dx}\left(\frac{dH}{dx}\right)=\frac{d}{dx}\left[180\left(1-\frac{4}{x^2}\right)\right]
\displaystyle =\frac{8}{x^3}\times180
\displaystyle \text{At }x=2,\ \left[\frac{d^2H}{dx^2}\right]_{x=2}=\frac{8}{2^3}\times180=180>0
\displaystyle \frac{d^2H}{dx^2}>0
\displaystyle \Rightarrow H\text{ is least at }x=2
\displaystyle \text{Also, the least cost}=280+180\left(2+\frac{4}{2}\right)
\displaystyle \text{[put }x=2\text{ in Eq. (iii) to get least cost }H\text{]}
\displaystyle =280+180\times4=280+720=\text{Rs.}1000
\displaystyle \text{Hence, the cost of least expensive tank is Rs. 1000.}
\\

\displaystyle \textbf{Question 34. }\text{Prove that the height of the cylinder of maximum volume that} \\ \text{can be inscribed in a sphere of radius }R\text{ is }\dfrac{2R}{\sqrt{3}}.\text{ Also, find the maximum volume.} \\ \hspace{0.2cm} \text{[CBSE 2019, 2014, 2012C, 2011; CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }h\text{ be the height and }a\text{ be the radius of base of cylinder inscribed in the} \\ \text{given sphere of radius }(R)
\displaystyle \text{In }\triangle ABC,\ AB^2+AC^2=BC^2\qquad\text{[by Pythagoras theorem]}
\displaystyle \Rightarrow a^2+\left(\frac{h}{2}\right)^2=R^2\Rightarrow a^2=R^2-\frac{h^2}{4}
\displaystyle \text{Volume of cylinder, }V=\pi a^2h
\displaystyle =\pi h\left(R^2-\frac{h^2}{4}\right)=\frac{\pi}{4}(4R^2h-h^3)
\displaystyle \text{On differentiating both sides two times w.r.t. }h\text{, we get}
\displaystyle \frac{dV}{dh}=\frac{\pi}{4}(4R^2-3h^2)
\displaystyle \text{and }\frac{d^2V}{dh^2}=\frac{\pi}{4}(-6h)=-\frac{3\pi h}{2}\qquad\ldots\text{(i)}
\displaystyle \text{For maxima or minima, put }\frac{dV}{dh}=0
\displaystyle \Rightarrow \frac{\pi}{4}(4R^2-3h^2)=0
\displaystyle \Rightarrow h^2=\frac{4}{3}R^2
\displaystyle \Rightarrow h=\frac{2}{\sqrt{3}}R
\displaystyle [\because\text{height is always positive, so we do not take }-\text{ve sign}]
\displaystyle \text{On substituting the value of }h\text{ in Eq. (i), we get}
\displaystyle \frac{d^2V}{dh^2}=-\frac{3\pi}{2}\cdot\frac{2}{\sqrt{3}}R=-\sqrt{3}\pi R<0
\displaystyle \Rightarrow V\text{ is maximum.}
\displaystyle \text{Hence, the required height of cylinder is }\frac{2R}{\sqrt{3}}\qquad\text{Hence proved.}
\displaystyle \text{Now, maximum volume of cylinder,}
\displaystyle V=\pi h\left(R^2-\frac{h^2}{4}\right)=\pi\cdot\frac{2R}{\sqrt{3}}\left(R^2-\frac{1}{4}\cdot\frac{4}{3}R^2\right)
\displaystyle \left[\because h=\frac{2}{\sqrt{3}}R\right]
\displaystyle =\frac{2\pi R}{\sqrt{3}}\cdot\frac{(3R^2-R^2)}{3}=\frac{4\pi R^3}{3\sqrt{3}}\text{ cu units}
\\

\displaystyle \textbf{Question 35. }\text{Find the point on the curve }y^2=4x\text{, which is nearest to the} \\ \text{point }(2,-8). \hspace{0.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation of curve is }y^2=4x
\displaystyle \text{Let }P(x,y)\text{ be a point on the curve, which is nearest to point }A(2,-8)
\displaystyle \text{Now, distance between the points }A\text{ and }P\text{ is given by}
\displaystyle AP=\sqrt{(x-2)^2+(y+8)^2}=\sqrt{\left(\frac{y^2}{4}-2\right)^2+(y+8)^2}
\displaystyle =\sqrt{\frac{y^4}{16}-y^2+4+y^2+16y+64}
\displaystyle =\sqrt{\frac{y^4}{16}+16y+68}
\displaystyle \text{Let }z=(AP)^2=\frac{y^4}{16}+16y+68
\displaystyle \text{Now, }\frac{dz}{dy}=\frac{1}{16}\times4y^3+16=\frac{y^3}{4}+16
\displaystyle \text{For maximum or minimum value of }z\text{, put}
\displaystyle \frac{dz}{dy}=0\Rightarrow \frac{y^3}{4}+16=0
\displaystyle \Rightarrow y^3+64=0\Rightarrow(y+4)(y^2-4y+16)=0
\displaystyle \Rightarrow y=-4
\displaystyle [\because y^2-4y+16=0\text{ gives imaginary values of }y]
\displaystyle \text{Now, }\frac{d^2z}{dy^2}=\frac{1}{4}\times3y^2=\frac{3}{4}y^2
\displaystyle \text{For }y=-4,
\displaystyle \frac{d^2z}{dy^2}=\frac{3}{4}(-4)^2=12>0
\displaystyle \text{Thus, }z\text{ is minimum when }y=-4
\displaystyle \text{Substituting }y=-4\text{ in equation of the curve }y^2=4x\text{; we obtain }x=4
\displaystyle \text{Hence, the point }(4,-4)\text{ on the curve }y^2=4x\text{ is nearest to the point }(2,-8)
\\

\displaystyle \textbf{Question 36. }\text{Show that the altitude of the right circular cone of maximum volume} \\ \text{that can be inscribed in a sphere of radius }r\text{ is }\dfrac{4r}{3}.\text{ Also, find the maximum} \\ \text{volume in terms of volume of the sphere.} \hspace{1.2cm} \text{[CBSE 2019, 2016]}
\displaystyle \textbf{Or}
\displaystyle \text{Show that the altitude of the right circular cone of maximum volume that can be inscribed} \\ \text{in a sphere of radius }r\text{ is }\dfrac{4r}{3}.\text{ Also, show that the maximum volume of the cone is }\dfrac{8}{27} \\ \text{of the volume of the sphere.} \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }R\text{ be the radius and }h\text{ be the height of the cone, which inscribed in a sphere of radius }r
\displaystyle \therefore OA=h-r
\displaystyle \text{In }\triangle OAB\text{, by Pythagoras theorem, we have}
\displaystyle r^2=R^2+(h-r)^2
\displaystyle \Rightarrow r^2=R^2+h^2+r^2-2rh
\displaystyle \Rightarrow R^2=2rh-h^2\qquad\ldots\text{(i)}
\displaystyle \text{The volume of sphere}=\frac{4}{3}\pi r^3
\displaystyle \text{and the volume }V\text{ of the cone,}
\displaystyle V=\frac{1}{3}\pi R^2h
\displaystyle \Rightarrow V=\frac{1}{3}\pi h(2rh-h^2)\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow V=\frac{1}{3}\pi(2rh^2-h^3)\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides of Eq. (ii) w.r.t. }h\text{, we get}
\displaystyle \frac{dV}{dh}=\frac{1}{3}\pi(4rh-3h^2)\qquad\ldots\text{(iii)}
\displaystyle \text{For maxima or minima, put }\frac{dV}{dh}=0
\displaystyle \Rightarrow \frac{1}{3}\pi(4rh-3h^2)=0
\displaystyle \Rightarrow 4rh=3h^2\Rightarrow 4r=3h
\displaystyle \Rightarrow h=\frac{4r}{3}\qquad[\because h\neq0]
\displaystyle \text{Again, on differentiating Eq. (iii) w.r.t. }h\text{, we get}
\displaystyle \frac{d^2V}{dh^2}=\frac{1}{3}\pi(4r-6h)
\displaystyle \text{At }h=\frac{4r}{3},\ \left(\frac{d^2V}{dh^2}\right)_{h=\frac{4r}{3}}=\frac{\pi}{3}\left(4r-6\times\frac{4r}{3}\right)
\displaystyle =\frac{\pi}{3}(4r-8r)=-\frac{4\pi r}{3}<0
\displaystyle \Rightarrow V\text{ is maximum at }h=\frac{4r}{3}\qquad\text{Hence proved.}
\displaystyle \text{On substituting the value of }h\text{ in Eq. (ii), we get}
\displaystyle V=\frac{1}{3}\pi\left[2r\left(\frac{4r}{3}\right)^2-\left(\frac{4r}{3}\right)^3\right]
\displaystyle =\frac{\pi}{3}\left[\frac{32}{9}r^3-\frac{64}{27}r^3\right]=\frac{\pi}{3}r^3\left[\frac{32}{9}-\frac{64}{27}\right]
\displaystyle =\frac{\pi}{3}r^3\left[\frac{96-64}{27}\right]=\frac{\pi}{3}r^3\left(\frac{32}{27}\right)
\displaystyle =\frac{8}{27}\times\left(\frac{4}{3}\pi r^3\right)=\frac{8}{27}\times(\text{volume of sphere})
\displaystyle \text{Hence, the maximum volume of the cone is }\frac{8}{27}\text{ of the volume of the sphere.}
\\

\displaystyle \textbf{Question 37. }\text{A window is of the form of a semi-circle with a rectangle on } \\ \text{its diameter. The total perimeter of the window is 10 m. Find the dimensions of the} \\ \text{window to admit maximum light through the whole opening.} \\  \hspace{1.2cm} \text{[CBSE 2018C; CBSE 2017, 2011; CBSE 2014]}
\displaystyle \textbf{Or}
\displaystyle \text{A window is in the form of a rectangle surmounted by a semicircular opening. The} \\ \text{total perimeter of the window is 10 m. Find the dimensions of the window to admit } \\ \text{maximum light through the whole opening.} \hspace{1.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }2x\text{ be the length and }y\text{ be the width of the window.}
\displaystyle \text{Then, radius of semicircular opening}=x\text{ m}
\displaystyle \text{Since, the perimeter of the window is 10 m.}
\displaystyle \therefore 2x+y+y+\frac{2\pi x}{2}=10
\displaystyle \Rightarrow 2x+2y+\pi x=10
\displaystyle \Rightarrow x(\pi+2)+2y=10
\displaystyle \Rightarrow y=\frac{10-x(\pi+2)}{2}\qquad\ldots\text{(i)}
\displaystyle \text{Note that to admit maximum light, area of window should be maximum.}
\displaystyle \text{Here, area of window}
\displaystyle A=\text{Area of rectangle}+\text{Area of semicircular region}
\displaystyle =2x\times y+\frac{1}{2}\pi x^2
\displaystyle \Rightarrow A=2x\left(\frac{10-x(\pi+2)}{2}\right)+\frac{1}{2}\pi x^2\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow A=10x-x^2(\pi+2)+\frac{1}{2}\pi x^2
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dA}{dx}=10-2x(\pi+2)+\pi x
\displaystyle =10-2\pi x-4x+\pi x
\displaystyle =10-\pi x-4x\qquad\ldots\text{(ii)}
\displaystyle \text{For maximum, put }\frac{dA}{dx}=0
\displaystyle \Rightarrow 10=\pi x+4x
\displaystyle \Rightarrow x=\frac{10}{\pi+4}
\displaystyle \text{Again, on differentiating both sides of Eq. (ii), we get}
\displaystyle \frac{d^2A}{dx^2}=-\pi-4
\displaystyle \Rightarrow \frac{d^2A}{dx^2}\bigg|_{x=\frac{10}{\pi+4}}=-(\pi+4)<0
\displaystyle \text{Thus, area is maximum when }x=\frac{10}{\pi+4}
\displaystyle \text{Now, on substituting the value of }x\text{ in Eq. (i), we get}
\displaystyle 2y=10-(\pi+2)\times\frac{10}{\pi+4}
\displaystyle \Rightarrow 2y=10\left[\frac{\pi+4-\pi-2}{\pi+4}\right]=\frac{20}{\pi+4}
\displaystyle \therefore y=\frac{10}{\pi+4}
\displaystyle \text{Hence, length of window}=\frac{20}{\pi+4}\text{ m and width of window}=\frac{10}{\pi+4}\text{ m, to admit maximum light through the whole opening.}
\\

\displaystyle \textbf{Question 38. }\text{Show that the surface area of a closed cuboid with square base and} \\ \text{given volume is minimum, when it is a cube.} \hspace{1.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }V\text{ be the fixed volume of a closed cuboid with length }x\text{, breadth }x\text{ and height }y
\displaystyle \text{Then, }V=x\times x\times y
\displaystyle \Rightarrow y=\frac{V}{x^2}\qquad\ldots\text{(i)}
\displaystyle \text{Let }S\text{ be its surface area.}
\displaystyle \text{Then, }S=2(x^2+xy+xy)
\displaystyle \Rightarrow S=2(x^2+2xy)
\displaystyle =2\left(x^2+\frac{2V}{x}\right)\qquad\text{[using Eq. (i)]}
\displaystyle \Rightarrow S=2\left(x^2+\frac{2V}{x}\right)\Rightarrow \frac{dS}{dx}=2\left(2x-\frac{2V}{x^2}\right)
\displaystyle \text{and }\frac{d^2S}{dx^2}=\left(4+\frac{8V}{x^3}\right)
\displaystyle \text{Now, }\frac{dS}{dx}=0\Rightarrow 2\left(2x-\frac{2V}{x^2}\right)=0
\displaystyle \Rightarrow 2x=\frac{2V}{x^2}
\displaystyle \Rightarrow x^3=V\Rightarrow V=x^3
\displaystyle \Rightarrow x\times x\times y=x^3
\displaystyle \Rightarrow y=x
\displaystyle \text{Also, }\left(\frac{d^2S}{dx^2}\right)_{x=y}=4+\frac{8V}{V}=12>0
\displaystyle \text{So, }S\text{ is minimum when length}=x\text{, breadth}=x\text{ and height}=x\text{, when it is cube.}\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 39. }\text{AB is the diameter of a circle and }C\text{ is any point on the circle. Show} \\ \text{that the area of }\triangle ABC\text{ is maximum, when it is an isosceles triangle.} \hspace{0.2cm} \text{[CBSE 2017, 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AC=x,\ BC=y\text{ and }r\text{ be the radius of circle.}
\displaystyle \text{Also, }\angle C=90^{\circ}\text{ [ }\because\text{angle made in semi-circle is }90^{\circ}\text{]}
\displaystyle \text{In }\triangle ABC\text{, we have}
\displaystyle AB^2=AC^2+BC^2
\displaystyle \Rightarrow (2r)^2=(x)^2+(y)^2
\displaystyle \Rightarrow 4r^2=x^2+y^2\qquad\ldots\text{(i)}
\displaystyle \text{We know that area of }\triangle ABC,\ A=\frac{1}{2}\times y
\displaystyle \text{On squaring both sides, we get}
\displaystyle A^2=\frac{1}{4}x^2y^2
\displaystyle \text{Let }A^2=S
\displaystyle \text{Then, }S=\frac{1}{4}x^2y^2
\displaystyle \Rightarrow S=\frac{1}{4}x^2(4r^2-x^2)\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow S=\frac{1}{4}(4r^2x^2-x^4)
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dS}{dx}=\frac{1}{4}(8r^2x-4x^3)
\displaystyle \text{For maxima or minima, put }\frac{dS}{dx}=0
\displaystyle \therefore \frac{1}{4}(8r^2x-4x^3)=0
\displaystyle \Rightarrow 8r^2x=4x^3
\displaystyle \Rightarrow 8r^2=4x^2\Rightarrow x^2=2r^2
\displaystyle \Rightarrow x=\sqrt{2}\,r
\displaystyle \text{From Eq. (i), we get}
\displaystyle y^2=4r^2-2r^2\Rightarrow y=\sqrt{2}\,r
\displaystyle \text{Here, }x=y\text{, so triangle is an isosceles.}
\displaystyle \text{Also, }\frac{d^2S}{dx^2}=\frac{d}{dx}\left[\frac{1}{4}(8r^2x-4x^3)\right]=\frac{1}{4}(8r^2-12x^2)
\displaystyle =2r^2-3x^2
\displaystyle \text{At }x=\sqrt{2}\,r,\ \frac{d^2S}{dx^2}=2r^2-3(2r^2)<0=-4r^2<0
\displaystyle \text{Hence, the area is maximum when triangle is an isosceles.}\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 40. }\text{A metal box with a square base and vertical sides is to contain} \\ \text{1024 cm}^3.\text{ The material for the top and bottom costs Rs. 5 per cm}^2\text{ and the} \\ \text{material for the sides costs Rs. 2.50 per cm}^2.\text{ Find the least cost of the box.} \hspace{0.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, volume of the box}=1024\text{ cm}^3.\text{ Let length of the side of square} \\ \text{base be }x\text{ cm and height of the box be }y\text{ cm.}
\displaystyle \text{Volume of the box, }V=x^2\cdot y=1024
\displaystyle \Rightarrow x^2y=1024\Rightarrow y=\frac{1024}{x^2}\qquad\ldots\text{(i)}
\displaystyle \text{Let }C\text{ denotes the cost of the box.}
\displaystyle C=2x^2\times5+4xy\times2.50
\displaystyle =10x^2+10xy
\displaystyle =10x\left(x+\frac{1024}{x^2}\right)
\displaystyle =10x^2+\frac{10240}{x}\qquad\ldots\text{(ii) [using Eq. (i)]}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dC}{dx}=20x-\frac{10240}{x^2}\qquad\ldots\text{(iii)}
\displaystyle \text{Now, }\frac{dC}{dx}=0
\displaystyle \Rightarrow 20x=\frac{10240}{x^2}
\displaystyle \Rightarrow 20x^3=10240
\displaystyle \Rightarrow x^3=512=8^3\Rightarrow x=8
\displaystyle \text{Again, on differentiating Eq. (iii) w.r.t. }x\text{, we get}
\displaystyle \frac{d^2C}{dx^2}=20+10240(-2)\cdot\frac{1}{x^3}
\displaystyle =20+\frac{20480}{x^3}>0
\displaystyle \left(\frac{d^2C}{dx^2}\right)_{x=8}=20+\frac{20480}{512}=60>0
\displaystyle \text{For }x=8\text{, cost is minimum and the corresponding least cost of the box}
\displaystyle C(8)=10\cdot8^2+\frac{10240}{8}=640+1280=1920
\displaystyle \text{[using Eq. (i)]}
\displaystyle \text{Hence, the least cost of the box is Rs. 1920.}
\\

\displaystyle \textbf{Question 41. }\text{Show that the semi-vertical angle of the cone of the maximum} \\ \text{volume and of given slant height is }\cos^{-1}\dfrac{1}{\sqrt{3}}. \hspace{1.2cm} \text{[CBSE 2016; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\theta\text{ be the semi-vertical angle of the cone.}
\displaystyle \text{It is clear that }\theta\in\left(0,\frac{\pi}{2}\right)
\displaystyle \text{Let }r,\ h\text{ and }l\text{ be the radius, height and the slant height of the cone, respectively.}
\displaystyle \text{Since, slant height of the cone is given, so consider it as constant.}
\displaystyle \text{Now, in }\triangle ABC,\ r=l\sin\theta\text{ and }h=l\cos\theta
\displaystyle \text{Let }V\text{ be the volume of the cone.}
\displaystyle \text{Then, }V=\frac{\pi}{3}r^2h\Rightarrow V=\frac{1}{3}\pi(l^2\sin^2\theta)(l\cos\theta)
\displaystyle \Rightarrow V=\frac{1}{3}\pi l^3\sin^2\theta\cos\theta
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{ two times, we get}
\displaystyle \frac{dV}{d\theta}=\frac{l^3\pi}{3}[\sin^2\theta(-\sin\theta)+\cos\theta(2\sin\theta\cos\theta)]
\displaystyle =\frac{l^3\pi}{3}(-\sin^3\theta+2\sin\theta\cos^2\theta)
\displaystyle \text{and }\frac{d^2V}{d\theta^2}=\frac{l^3\pi}{3}(-3\sin^2\theta\cos\theta+2\cos^3\theta-4\sin^2\theta\cos\theta)
\displaystyle \Rightarrow \frac{d^2V}{d\theta^2}=\frac{l^3\pi}{3}(2\cos^3\theta-7\sin^2\theta\cos\theta)
\displaystyle \text{For maxima or minima, put }\frac{dV}{d\theta}=0
\displaystyle \Rightarrow \sin^3\theta=2\sin\theta\cos^2\theta\Rightarrow\tan^2\theta=2
\displaystyle \Rightarrow \tan\theta=\sqrt{2}\Rightarrow\theta=\tan^{-1}\sqrt{2}
\displaystyle \text{Now, when }\theta=\tan^{-1}\sqrt{2}\text{, then }\tan^2\theta=2
\displaystyle \Rightarrow \sin^2\theta=2\cos^2\theta
\displaystyle \text{Now, we have}
\displaystyle \frac{d^2V}{d\theta^2}=\frac{l^3\pi}{3}(2\cos^3\theta-14\cos^3\theta)
\displaystyle =-4\pi l^3\cos^3\theta<0,\text{ for }\theta\in\left(0,\frac{\pi}{2}\right)
\displaystyle \therefore V\text{ is maximum, when }\theta=\tan^{-1}\sqrt{2}
\displaystyle \text{or }\theta=\cos^{-1}\frac{1}{\sqrt{3}}
\displaystyle \left[\because\cos\theta=\frac{1}{\sqrt{1+\tan^2\theta}}=\frac{1}{\sqrt{1+2}}=\frac{1}{\sqrt{3}}\right]
\displaystyle \text{Hence, for given slant height, the semi-vertical angle of the cone of maximum volume is }\cos^{-1}\frac{1}{\sqrt{3}}\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 42. }\text{Prove that the least perimeter of an isosceles triangle in} \\ \text{which a circle of radius }r\text{ can be inscribed, is }6\sqrt{3}\,r. \hspace{1.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABC\text{ be the given isosceles triangle, with }AB=AC
\displaystyle \text{Clearly, }OD\perp BC,\ OF\perp AC,\ OE\perp AB
\displaystyle [\because\text{radius is perpendicular to the tangent at the point of contact}]
\displaystyle \text{and }BD=BE,\ CD=CF,\ AE=AF
\displaystyle [\because\text{tangents from an external point to a circle are equal in length}]
\displaystyle \text{Since, }AD\text{ is an altitude of isosceles }\triangle ABC
\displaystyle \text{Therefore, }BD=CD\qquad[\because\text{in an isosceles triangle, altitude from common vertex of equal sides bisect the third side}]
\displaystyle \Rightarrow BD=BE=CF=CD\qquad[\because BE=BD\text{ and }CD=CF]
\displaystyle \text{Now, perimeter }(P)\text{ of }\triangle ABC=AB+BC+AC
\displaystyle =AE+BE+BD+DC+AF+FC
\displaystyle =(AE+AF)+(BE+BD+DC+FC)
\displaystyle =2AE+4BD\qquad\ldots\text{(i)}
\displaystyle \text{Consider }\triangle OEA\text{, we have}
\displaystyle AE=\frac{OE}{\tan\theta}=\frac{r}{\tan\theta}\text{ and }OA=\frac{r}{\sin\theta}
\displaystyle \text{and in }\triangle ADB\text{, we have }BD=AD\tan\theta
\displaystyle BD=(AO+OD)\tan\theta
\displaystyle =\left(\frac{r}{\sin\theta}+r\right)\tan\theta
\displaystyle \text{Now, }P=2\cdot\frac{r}{\tan\theta}+4\cdot\left(\frac{r}{\sin\theta}+r\right)\tan\theta\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow P(\theta)=r(2\cot\theta+4\sec\theta+4\tan\theta)\qquad\ldots\text{(ii)}
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{, we get}
\displaystyle P'(\theta)=r(-2\text{cosec}^2\theta+4\sec\theta\tan\theta+4\sec^2\theta)
\displaystyle \ldots\text{(iii)}
\displaystyle =r\left(\frac{-2}{\sin^2\theta}+\frac{4\sin\theta}{\cos^2\theta}+\frac{4}{\cos^2\theta}\right)
\displaystyle =r\left(\frac{-2\cos^2\theta+4\sin^3\theta+4\sin^2\theta}{\sin^2\theta\cos^2\theta}\right)
\displaystyle \text{Now, put }P'(\theta)=0
\displaystyle \Rightarrow -2\cos^2\theta+4\sin^3\theta+4\sin^2\theta=0
\displaystyle \Rightarrow -2(1-\sin^2\theta)+4\sin^3\theta+4\sin^2\theta=0
\displaystyle \Rightarrow -2+2\sin^2\theta+4\sin^3\theta+4\sin^2\theta=0
\displaystyle \Rightarrow 2\sin^3\theta+3\sin^2\theta-1=0
\displaystyle \Rightarrow (\sin\theta+1)(2\sin^2\theta+\sin\theta-1)=0
\displaystyle \Rightarrow \sin\theta=-1\text{ or }2\sin^2\theta+\sin\theta-1=0
\displaystyle \Rightarrow 2\sin^2\theta+\sin\theta-1=0
\displaystyle [\because\sin\theta\neq-1,\text{ as }\theta\text{ can't be more than }90^{\circ}]
\displaystyle \Rightarrow (2\sin\theta-1)(\sin\theta+1)=0
\displaystyle \Rightarrow \sin\theta=\frac{1}{2}\qquad[\because\sin\theta\neq-1]
\displaystyle \therefore \theta=\frac{\pi}{6}
\displaystyle \text{On differentiating both sides of Eq. (iii) w.r.t. }\theta\text{, we get}
\displaystyle P''(\theta)=r(4\text{cosec}^2\theta\cot\theta+4\sec^3\theta+4\sec\theta\tan^2\theta+8\sec^2\theta\tan\theta)
\displaystyle \Rightarrow P''\left(\frac{\pi}{6}\right)>0
\displaystyle \text{Thus, }P(\theta)\text{ is minimum, when }\theta=\frac{\pi}{6}
\displaystyle \text{Now, from Eq. (ii), least perimeter}=P\left(\frac{\pi}{6}\right)
\displaystyle =r\left[2\cot\left(\frac{\pi}{6}\right)+4\sec\left(\frac{\pi}{6}\right)+4\tan\left(\frac{\pi}{6}\right)\right]
\displaystyle =r\left[2\sqrt{3}+4\cdot\frac{2}{\sqrt{3}}+4\cdot\frac{1}{\sqrt{3}}\right]=r\left(\frac{6+8+4}{\sqrt{3}}\right)
\displaystyle =\frac{18}{\sqrt{3}}r=6\sqrt{3}\,r\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 43. }\text{Find the coordinates of a point on the parabola} \\ y=x^2+7x+2\text{ which is closest to the straight line }y=3x-3. \hspace{0.2cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, equation of curve is }y=x^2+7x+2\text{ and equation of straight line is }y=3x-3.
\displaystyle \text{Let }P(x,y)\text{ be any point on the parabola }y=x^2+7x+2.
\displaystyle \text{Let }D\text{ be the distance of point }P\text{ from straight line, then}
\displaystyle D=\frac{|3x-y-3|}{\sqrt{3^2+(-1)^2}}=\frac{|3x-y-3|}{\sqrt{9+1}}
\displaystyle =\frac{|3x-y-3|}{\sqrt{10}}=\frac{|3x-(x^2+7x+2)-3|}{\sqrt{10}}
\displaystyle =\frac{|{-}(x^2+4x+5)|}{\sqrt{10}}=\frac{x^2+2\cdot 2x+2^2+1}{\sqrt{10}}
\displaystyle \Rightarrow D=\frac{(x+2)^2+1}{\sqrt{10}}\qquad\ldots\text{(i)}
\displaystyle \text{On differentiating both sides of Eq. (i) w.r.t. }x\text{, we get}
\displaystyle \frac{dD}{dx}=\frac{2(x+2)+0}{\sqrt{10}}
\displaystyle \text{For extremum value of }D\text{, put }\frac{dD}{dx}=0
\displaystyle \Rightarrow 2(x+2)=0
\displaystyle \Rightarrow x=-2
\displaystyle \text{Now, }\frac{d^2D}{dx^2}=\frac{2}{\sqrt{10}}>0
\displaystyle \text{Thus, }D\text{ is minimum when }x=-2.
\displaystyle \text{Now, }y=x^2+7x+2=(-2)^2+7(-2)+2
\displaystyle =4-14+2=-8
\displaystyle \text{Hence, the point }(-2,-8)\text{ is on the parabola, which is closest to the given straight line.}
\\

\displaystyle \textbf{Question 44. }\text{A point on the hypotenuse of a right triangle is at distances }a\text{ and }b \\ \text{from the sides of the triangle. Show that the minimum length of the hypotenuse is } \\ (a^{2/3}+b^{2/3})^{3/2}. \hspace{1.2cm} \text{[CBSE 2015C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P\text{ be a point on the hypotenuse }AC\text{ of right angled }\triangle ABC\text{. Such that }PL\perp AB\text{ and }PL=a\text{ and }PM\perp BC\text{ and }PM=b.
\displaystyle \text{Let }\angle APL=\angle ACB=\theta\qquad\text{[say]}
\displaystyle \text{Then, }AP=a\sec\theta,\ PC=b\text{ cosec }\theta
\displaystyle \text{Let }l\text{ be the length of the hypotenuse, then}
\displaystyle l=AP+PC
\displaystyle \Rightarrow l=a\sec\theta+b\text{ cosec }\theta,\ 0<\theta<\frac{\pi}{2}
\displaystyle \text{On differentiating both sides w.r.t. }\theta\text{, we get}
\displaystyle \frac{dl}{d\theta}=a\sec\theta\tan\theta-b\text{ cosec }\theta\cot\theta\qquad\ldots\text{(i)}
\displaystyle \text{For maxima or minima, put }\frac{dl}{d\theta}=0
\displaystyle \Rightarrow a\sec\theta\tan\theta=b\text{ cosec }\theta\cot\theta
\displaystyle \Rightarrow \frac{a\sin\theta}{\cos^2\theta}=\frac{b\cos\theta}{\sin^2\theta}\Rightarrow\tan^3\theta=\left(\frac{b}{a}\right)
\displaystyle \Rightarrow\tan\theta=\left(\frac{b}{a}\right)^{1/3}
\displaystyle \text{Again, on differentiating both sides of Eq. (i) w.r.t. }\theta\text{, we get}
\displaystyle \frac{d^2l}{d\theta^2}=a(\sec\theta\times\sec^2\theta+\tan\theta\times\sec\theta\tan\theta)
\displaystyle -b[\text{cosec }\theta(-\text{cosec}^2\theta)+\cot\theta(-\text{cosec }\theta\cot\theta)]
\displaystyle =a\sec\theta(\sec^2\theta+\tan^2\theta)
\displaystyle +b\text{ cosec }\theta(\text{cosec}^2\theta+\cot^2\theta)
\displaystyle \text{For }0<\theta<\frac{\pi}{2}\text{, all trigonometric ratios are positive.}
\displaystyle \text{Also, }a>0\text{ and }b>0.
\displaystyle \therefore\ \frac{d^2l}{d\theta^2}\text{ is positive.}
\displaystyle \Rightarrow l\text{ is least when }\tan\theta=\left(\frac{b}{a}\right)^{\frac{1}{3}}
\displaystyle \therefore\text{ Least value of,}
\displaystyle l=a\sec\theta+b\text{ cosec }\theta
\displaystyle =a\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}+b\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}
\displaystyle =\sqrt{a^{2/3}+b^{2/3}}(a^{2/3}+b^{2/3})=(a^{2/3}+b^{2/3})^{3/2}
\displaystyle \left[\because\text{ in }\triangle EFG,\ \tan\theta=\frac{b^{1/3}}{a^{1/3}},\ \sec\theta=\frac{\sqrt{a^{2/3}+b^{2/3}}}{a^{1/3}}\right.
\displaystyle \left.\text{and cosec }\theta=\frac{\sqrt{a^{2/3}+b^{2/3}}}{b^{1/3}}\right]
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 45. }\text{The sum of the perimeters of a circle and square is }k\text{, where } \\ k\text{ is some constant. Prove that the sum of their areas is least, when the side of the} \\ \text{square is double the radius of the circle.} \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ be the radius of circle and }x\text{ be the side of a square. Then, given that}
\displaystyle \text{Perimeter of square}+\text{Perimeter of circle}=k\text{ (constant)}
\displaystyle \text{i.e. }4x+2\pi r=k
\displaystyle \Rightarrow x=\frac{k-2\pi r}{4}\qquad\ldots\text{(i)}
\displaystyle \text{Let }A\text{ denotes the sum of their areas.}
\displaystyle \therefore\text{Area, }A=\text{Area of a square}+\text{Area of circle}
\displaystyle \therefore A=x^2+\pi r^2\qquad\ldots\text{(ii)}
\displaystyle \text{On putting the value of }x\text{ from Eq. (i) in Eq. (ii), we get}
\displaystyle A=\left(\frac{k-2\pi r}{4}\right)^2+\pi r^2
\displaystyle \text{On differentiating both sides w.r.t. }r\text{, we get}
\displaystyle \frac{dA}{dr}=2\left(\frac{k-2\pi r}{4}\right)\left(\frac{-2\pi}{4}\right)+2\pi r
\displaystyle =-\frac{\pi}{4}(k-2\pi r)+2\pi r
\displaystyle \text{For maxima and minima, put }\frac{dA}{dr}=0
\displaystyle \Rightarrow -\frac{\pi}{4}(k-2\pi r)+2\pi r=0
\displaystyle \Rightarrow -\frac{\pi}{4}k+\frac{\pi^2r}{2}+2\pi r=0
\displaystyle \Rightarrow \frac{r\pi}{2}(\pi+4)=\frac{\pi}{4}k
\displaystyle \Rightarrow r=\frac{k}{2\pi+8}\qquad\ldots\text{(iii)}
\displaystyle \text{Now, }\frac{d^2A}{dr^2}=\frac{d}{dr}\left(\frac{dA}{dr}\right)=\frac{d}{dr}\left[2\pi r-\frac{\pi}{4}(k-2\pi r)\right]
\displaystyle =2\pi+\frac{2\pi^2}{4}=2\pi+\frac{\pi^2}{2}>0
\displaystyle \therefore \frac{d^2A}{dr^2}>0\text{, so }A\text{ is minimum.}
\displaystyle \text{From Eq. (iii), we get}
\displaystyle r=\frac{k}{2\pi+8}
\displaystyle \Rightarrow 2r\pi+8r=k
\displaystyle \Rightarrow 2r\pi+8r=4x+2\pi r\qquad[\because k=4x+2\pi r]
\displaystyle \therefore 8r=4x
\displaystyle \Rightarrow x=2r
\displaystyle \text{i.e. Side of square}=\text{Double the radius of circle}
\displaystyle \text{Hence, the sum of area of a circle and a square is least, when side of square is equal} \\ \text{to diameter of circle or double the radius of circle.}\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 46. }\text{Find the point }P\text{ on the curve }y^2=4ax\text{, which is nearest to} \\ \text{the point }(11a,0). \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }P(x,y)\text{ be any point on }y^2=4ax\qquad\ldots\text{(i)}
\displaystyle \text{Then, distance between }(x,y)\text{ and }(11a,0)\text{ is given by}
\displaystyle D=\sqrt{(x-11a)^2+(y-0)^2}
\displaystyle =\sqrt{(x-11a)^2+y^2}
\displaystyle =\sqrt{(x-11a)^2+4ax}\qquad\text{[from Eq. (i)]}
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle \frac{dD}{dx}=\frac{[2(x-11a)+4a]}{2\sqrt{(x-11a)^2+4ax}}
\displaystyle \Rightarrow\frac{dD}{dx}=\frac{2x-22a+4a}{2\sqrt{(x-11a)^2+4ax}}
\displaystyle \Rightarrow\frac{dD}{dx}=\frac{x-9a}{\sqrt{(x-11a)^2+4ax}}
\displaystyle \text{Put }\frac{dD}{dx}=0\Rightarrow x-9a=0\Rightarrow x=9a
\displaystyle \text{Now, differentiating both sides of }\frac{dD}{dx}\text{, we get}
\displaystyle \frac{d^2D}{dx^2}=\frac{d}{dx}\left(\frac{dD}{dx}\right)
\displaystyle =\frac{d}{dx}\left(\frac{x-9a}{\sqrt{(x-11a)^2+4ax}}\right)
\displaystyle =\frac{\sqrt{(x-11a)^2+4ax}-(x-9a)\cdot\dfrac{1\cdot[2(x-11a)+4a]}{2\sqrt{(x-11a)^2+4ax}}}{(x-11a)^2+4ax}
\displaystyle \therefore\left(\frac{d^2D}{dx^2}\right)_{x=9a}=\frac{\sqrt{(9a-11a)^2+4a\cdot 9a}-0}{(9a-11a)^2+4a\times 9a}
\displaystyle =\frac{\sqrt{4a^2+36a^2}}{4a^2+36a^2}=\frac{1}{\sqrt{40a^2}}>0
\displaystyle \text{So, at }(x=9a),\ D\text{ is minimum.}
\displaystyle \text{Now, }y^2=36a^2\Rightarrow y=\pm 6a
\displaystyle \text{Hence, the required points are }(9a,6a)\text{ and }(9a,-6a).
\\

\displaystyle \textbf{Question 47. }\text{Prove that the semi-vertical angle of the right circular cone of given} \\ \text{volume and least curved surface area is }\cot^{-1}\sqrt{2}. \hspace{0.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ be the radius of the base, }h\text{ be the height, }V\text{ be the volume, }\\ S\text{ be the surface area of the cone }ABC\text{ and }\theta\text{ be the semi-vertical angle.}
\displaystyle \text{Then, }V=\frac{1}{3}\pi r^2h\Rightarrow 3V=\pi r^2h
\displaystyle \Rightarrow 9V^2=\pi^2r^4h^2\qquad\text{[squaring both sides]}
\displaystyle \Rightarrow h^2=\frac{9V^2}{\pi^2r^4}\qquad\ldots\text{(i)}
\displaystyle \text{and curved surface area, }S=\pi rl
\displaystyle \Rightarrow S=\pi r\sqrt{r^2+h^2}\qquad\left[\because l=\sqrt{h^2+r^2}\right]
\displaystyle \Rightarrow S^2=\pi^2r^2(r^2+h^2)\qquad\text{[on squaring both sides]}
\displaystyle \Rightarrow S^2=\pi^2r^2\left(\frac{9V^2}{\pi^2r^4}+r^2\right)\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow S^2=\frac{9V^2}{r^2}+\pi^2r^4\qquad\ldots\text{(ii)}
\displaystyle \text{When }S\text{ is least, then }S^2\text{ is also least.}
\displaystyle \text{Now, }\frac{d}{dr}(S^2)=-\frac{18V^2}{r^3}+4\pi^2r^3\qquad\ldots\text{(iii)}
\displaystyle \text{For maxima or minima, put }\frac{d}{dr}(S^2)=0
\displaystyle \Rightarrow -\frac{18V^2}{r^3}+4\pi^2r^3=0
\displaystyle \Rightarrow 18V^2=4\pi^2r^6
\displaystyle \Rightarrow 9V^2=2\pi^2r^6\qquad\ldots\text{(iv)}
\displaystyle \text{Again, on differentiating Eq. (iii) w.r.t. }r\text{, we get}
\displaystyle \frac{d^2}{dr^2}(S^2)=\frac{54V^2}{r^4}+12\pi^2r^2>0
\displaystyle \text{At }r=\left(\frac{9V^2}{2\pi^2}\right)^{1/6},\ \frac{d^2}{dr^2}(S^2)>0
\displaystyle \text{So, }S^2\text{ or }S\text{ is minimum, when }V^2=2\pi^2r^6/9
\displaystyle \text{On putting }V^2=2\pi^2r^6/9\text{ in Eq. (i), we get}
\displaystyle 2\pi^2r^6=\pi^2r^4h^2
\displaystyle \Rightarrow 2r^2=h^2\Rightarrow h=\sqrt{2}\,r\Rightarrow\frac{h}{r}=\sqrt{2}
\displaystyle \Rightarrow\cot\theta=\sqrt{2}\qquad\left[\text{from the figure, }\cot\theta=\frac{h}{r}\right]
\displaystyle \therefore\ \theta=\cot^{-1}\sqrt{2}
\displaystyle \text{Hence, the semi-vertical angle of the right circular cone of given volume and least curved surface area is }\cot^{-1}\sqrt{2}.\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 48. }\text{Of all the closed right circular cylindrical cans of volume} \\ 128\pi\text{ cm}^3\text{, find the dimensions of the can which has minimum surface area.} \hspace{0.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ cm be the radius of base and }h\text{ cm be the height of the cylindrical can.} \\ \text{Let its volume be }V\text{ and }S\text{ be its total surface area.}
\displaystyle \text{Then, }V=128\pi\text{ cm}^3\qquad\text{[given]}
\displaystyle \Rightarrow \pi r^2h=128\pi\Rightarrow h=\frac{128}{r^2}\qquad\ldots\text{(i)}
\displaystyle \text{Now, surface area of cylindrical can,}
\displaystyle S=2\pi r^2+2\pi rh\qquad\ldots\text{(ii)}
\displaystyle \Rightarrow S=2\pi r^2+2\pi r\left(\frac{128}{r^2}\right)\qquad\text{[using Eq. (i)]}
\displaystyle \Rightarrow S=2\pi r^2+\frac{256\pi}{r}\qquad\ldots\text{(iii)}
\displaystyle \text{On differentiating both sides of Eq. (iii) w.r.t. }r\text{, we get}
\displaystyle \frac{dS}{dr}=4\pi r-\frac{256\pi}{r^2}\qquad\ldots\text{(iv)}
\displaystyle \text{For maxima or minima, put }\frac{dS}{dr}=0
\displaystyle \Rightarrow 4\pi r=\frac{256\pi}{r^2}
\displaystyle \Rightarrow r^3=\frac{256}{4}\Rightarrow r^3=64
\displaystyle \text{Taking cube root on both sides, we get}
\displaystyle r=(64)^{1/3}
\displaystyle \Rightarrow r=4\text{ cm}
\displaystyle \text{Again, on differentiating Eq. (iv) w.r.t. }r\text{, we get}
\displaystyle \frac{d^2S}{dr^2}=4\pi+\frac{512\pi}{r^3}
\displaystyle \text{At }r=4,
\displaystyle \left(\frac{d^2S}{dr^2}\right)_{r=4}=\frac{512\pi}{64}+4\pi=8\pi+4\pi=12\pi>0
\displaystyle \text{Thus, }\frac{d^2S}{dr^2}>0\text{ at }r=4\text{, so the surface area is minimum, when the radius of cylinder is 4 cm.}
\displaystyle \text{On putting the value of }r\text{ in Eq. (i), we get}
\displaystyle h=\frac{128}{(4)^2}=\frac{128}{16}=8\text{ cm}
\displaystyle \text{Hence, for the minimum surface area of can, the dimensions of the cylindrical can are }r=4\text{ cm and }h=8\text{ cm.}
\\

\displaystyle \textbf{Question 49. }\text{Show that a cylinder of a given volume which is open at the top has} \\ \text{minimum total surface area, when its height is equal to the radius of its base.} \\ \hspace{0.2cm} \text{[CBSE 2014; CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }r\text{ be the radius, }h\text{ be the height, }V\text{ be the volume and }S\text{ be the total surface area of a right circular cylinder which is open at the top.}
\displaystyle \text{Now, given that }V=\pi r^2h
\displaystyle \Rightarrow h=\frac{V}{\pi r^2}\qquad\ldots\text{(i)}
\displaystyle \text{We know that total surface area }S\text{ is given by}
\displaystyle S=2\pi rh+\pi r^2
\displaystyle [\because\text{ cylinder is open at the top, therefore }S=\text{Curved surface area of cylinder}+\text{Area of base}]
\displaystyle \Rightarrow S=2\pi r\left(\frac{V}{\pi r^2}\right)+\pi r^2\qquad\left[\text{put }h=\frac{V}{\pi r^2}\text{, from Eq. (i)}\right]
\displaystyle \Rightarrow S=\frac{2V}{r}+\pi r^2
\displaystyle \text{On differentiating both sides w.r.t. }r\text{, we get}
\displaystyle \frac{dS}{dr}=-\frac{2V}{r^2}+2\pi r
\displaystyle \text{For maxima or minima, put }\frac{dV}{dr}=0
\displaystyle \Rightarrow -\frac{2V}{r^2}+2\pi r=0\Rightarrow V=\pi r^3
\displaystyle \Rightarrow \pi r^2h=\pi r^3\qquad[\because V=\pi r^2h]
\displaystyle \Rightarrow h=r
\displaystyle \text{Also, }\frac{d^2S}{dr^2}=\frac{d}{dr}\left(\frac{dS}{dr}\right)
\displaystyle =\frac{d}{dr}\left(\frac{-2V}{r^2}+2\pi r\right)
\displaystyle \Rightarrow \frac{d^2S}{dr^2}=\frac{4V}{r^3}+2\pi
\displaystyle \text{On putting }r=h\text{, we get}
\displaystyle \left[\frac{d^2S}{dr^2}\right]_{r=h}=\frac{4V}{h^3}+2\pi>0,\text{ as }h>0.
\displaystyle \text{Then, }\frac{d^2S}{dr^2}>0\Rightarrow S\text{ is minimum.}
\displaystyle \text{Hence, }S\text{ is minimum, when }h=r\text{ i.e. when height of cylinder is equal to radius of the base.}\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 50. }\text{Prove that the area of a right angled triangle of given hypotenuse is} \\ \text{maximum, when the triangle is isosceles.} \hspace{1.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ and }b\text{ be the sides of right angled triangle and }c\text{ be the hypotenuse.}
\displaystyle \text{From }\triangle ABC\text{, we have }c^2=a^2+b^2
\displaystyle \text{Area of }\triangle ABC,\ (A)=\frac{1}{2}\cdot a\cdot b=\frac{1}{2}a\sqrt{c^2-a^2}
\displaystyle [\because b=\sqrt{c^2-a^2}]
\displaystyle \text{On differentiating both sides w.r.t. }a\text{, we get}
\displaystyle \frac{dA}{da}=\frac{1}{2}\cdot 1\cdot\sqrt{c^2-a^2}+\frac{1}{2}\cdot a\cdot\frac{(-2a)}{2\sqrt{c^2-a^2}}
\displaystyle =\frac{1}{2}\left(\sqrt{c^2-a^2}-\frac{a^2}{\sqrt{c^2-a^2}}\right)
\displaystyle \text{For maxima or minima, put }\frac{dA}{da}=0
\displaystyle \Rightarrow \frac{1}{2}\left(\sqrt{c^2-a^2}-\frac{a^2}{\sqrt{c^2-a^2}}\right)=0\Rightarrow c^2-a^2-a^2=0
\displaystyle \Rightarrow c^2=2a^2\Rightarrow a=\frac{c}{\sqrt{2}}
\displaystyle \text{Now, }\frac{d^2A}{da^2}=\frac{1}{2}\left[\frac{-a}{\sqrt{c^2-a^2}}-\frac{a^3}{(c^2-a^2)^{3/2}}\right]
\displaystyle =-\frac{1}{2}a\left[\frac{c^2-a^2+a^2}{(c^2-a^2)^{3/2}}\right]
\displaystyle =-\frac{1}{2}\cdot\frac{c^2a}{(c^2-a^2)^{3/2}}<0
\displaystyle \therefore\text{ Area of }\triangle ABC\text{ is maximum and}
\displaystyle b=\sqrt{c^2-a^2}=\sqrt{2a^2-a^2}=a
\displaystyle \text{Hence, the triangle is isosceles.}\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 51. }\text{Show that the right circular cone of least curved surface and given} \\ \text{volume has an altitude equal to }\sqrt{2}\text{ times the radius of the base.} \hspace{0.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }C\text{ denotes the curved surface area, }r\text{ be the radius of base, }h\text{ be the height and } \\ V\text{ be the volume of right circular cone.}
\displaystyle \text{We know that volume of cone is given by}
\displaystyle V=\frac{1}{3}\pi r^2h\Rightarrow h=\frac{3V}{\pi r^2}\qquad\ldots\text{(i)}
\displaystyle \text{Also, the curved surface area of cone is given by}
\displaystyle C=\pi rl\text{, where }l=\sqrt{r^2+h^2}\text{ is the slant height of cone.}
\displaystyle \therefore\ C=\pi r\sqrt{r^2+h^2}
\displaystyle \text{On squaring both sides, we get}
\displaystyle C^2=\pi^2r^2(r^2+h^2)\Rightarrow C^2=\pi^2r^4+\pi^2r^2h^2
\displaystyle \text{Let }C^2=Z
\displaystyle \text{Then, }Z=\pi^2r^4+\pi^2r^2h^2\qquad\ldots\text{(ii)}
\displaystyle \Rightarrow Z=\pi^2r^4+\pi^2r^2\left(\frac{3V}{\pi r^2}\right)^2\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow Z=\pi^2r^4+\pi^2r^2\times\frac{9V^2}{\pi^2r^4}
\displaystyle \Rightarrow Z=\pi^2r^4+\frac{9V^2}{r^2}
\displaystyle \text{On differentiating both sides w.r.t. }r\text{, we get}
\displaystyle \frac{dZ}{dr}=4\pi^2r^3-\frac{18V^2}{r^3}
\displaystyle \text{For maxima or minima, put }\frac{dZ}{dr}=0
\displaystyle \Rightarrow 4\pi^2r^3-\frac{18V^2}{r^3}=0\Rightarrow 4\pi^2r^3=\frac{18V^2}{r^3}
\displaystyle \Rightarrow 4\pi^2r^6=18\left(\frac{1}{3}\pi r^2h\right)^2\qquad\left[\because V=\frac{1}{3}\pi r^2h\right]
\displaystyle \Rightarrow 4\pi^2r^6=18\times\frac{1}{9}\pi^2r^4h^2
\displaystyle \Rightarrow 4\pi^2r^6=2\pi^2r^4h^2\Rightarrow 2r^2=h^2
\displaystyle \therefore\ h=\sqrt{2}\,r
\displaystyle \text{Hence, height}=\sqrt{2}\times\text{(radius of base)}
\displaystyle \text{Also, }\frac{d^2Z}{dr^2}=\frac{d}{dr}\left(\frac{dZ}{dr}\right)=\frac{d}{dr}\left(4\pi^2r^3-\frac{18V^2}{r^3}\right)
\displaystyle =12\pi^2r^2+\frac{54V^2}{r^4}
\displaystyle \therefore\ \frac{d^2Z}{dr^2}=12\pi^2r^2+\frac{54V^2}{r^4}>0
\displaystyle \Rightarrow Z\text{ is minimum}\Rightarrow C\text{ is minimum.}
\displaystyle \text{Hence, the curved surface area is least, when }h=\sqrt{2}\,r.\qquad\text{Hence proved.}
\\

\displaystyle \textbf{Question 52. }\text{A window has the shape of a rectangle surmounted by an equilateral} \\ \text{triangle. If the perimeter of the window is 12 m, then find the dimensions of the rectangle} \\ \text{that will produce the largest area of the window.} \hspace{1.2cm} \text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABCD\text{ be the rectangle which is surmounted by an equilateral }\triangle EDC.
\displaystyle \text{Now, given that Perimeter of window}=12\text{ m}\Rightarrow 2x+2y+y=12
\displaystyle \therefore\ x=6-\frac{3}{2}y\qquad\ldots\text{(i)}
\displaystyle \text{Let }A\text{ denotes the combined area of the window.}
\displaystyle \text{Then, }A=\text{area of rectangle}+\text{area of equilateral triangle}
\displaystyle \Rightarrow A=xy+\frac{\sqrt{3}}{4}y^2
\displaystyle \Rightarrow A=y\left(6-\frac{3}{2}y\right)+\frac{\sqrt{3}}{4}y^2
\displaystyle \left[\because x=6-\frac{3}{2}y\text{ from Eq. (i)}\right]
\displaystyle \Rightarrow A=6y-\frac{3}{2}y^2+\frac{\sqrt{3}}{4}y^2
\displaystyle \text{On differentiating both sides w.r.t. }y\text{, we get}
\displaystyle \frac{dA}{dy}=6-3y+\frac{\sqrt{3}}{2}y
\displaystyle \text{For maxima or minima, put }\frac{dA}{dy}=0
\displaystyle \Rightarrow 6-3y+\frac{\sqrt{3}}{2}y=0\Rightarrow y\left(\frac{\sqrt{3}}{2}-3\right)=-6
\displaystyle \Rightarrow y=\frac{12}{6-\sqrt{3}}
\displaystyle \text{Now, }\frac{d^2A}{dy^2}=\frac{d}{dy}\left(\frac{dA}{dy}\right)=\frac{d}{dy}\left(6-3y+\frac{\sqrt{3}}{2}y\right)
\displaystyle =-3+\frac{\sqrt{3}}{2}=\frac{-6+\sqrt{3}}{2}<0
\displaystyle \therefore A\text{ is maximum.}
\displaystyle \text{Now, on putting }y=\frac{12}{6-\sqrt{3}}\text{ in Eq. (i), we get}
\displaystyle x=6-\frac{3}{2}\left(\frac{12}{6-\sqrt{3}}\right)\Rightarrow x=\frac{36-6\sqrt{3}-18}{6-\sqrt{3}}
\displaystyle \therefore\ x=\frac{18-6\sqrt{3}}{6-\sqrt{3}}
\displaystyle \text{Hence, the area of the window is largest when the dimensions of the window are}
\displaystyle x=\frac{18-6\sqrt{3}}{6-\sqrt{3}}\text{ and }y=\frac{12}{6-\sqrt{3}}
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.