\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}.

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