Class 12: Maxima and Minima

$\displaystyle \textbf{Question 1: }~\text{Determine two positive numbers whose sum is }15\text{ and the sum } \\ \text{of whose squares is minimum.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the two positive numbers be } x \text{ and } y.$

$\displaystyle x+y=15 \qquad \text{(1)}$

$\displaystyle \text{Now,}$

$\displaystyle z=x^{2}+y^{2}$

$\displaystyle \Rightarrow z=x^{2}+(15-x)^{2} \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow z=x^{2}+x^{2}+225-30x$

$\displaystyle \Rightarrow z=2x^{2}-30x+225$

$\displaystyle \Rightarrow \frac{dz}{dx}=4x-30$

$\displaystyle \text{For maximum or minimum values of } z, \text{ we must have } \frac{dz}{dx}=0$

$\displaystyle \Rightarrow 4x-30=0$

$\displaystyle \Rightarrow x=\frac{15}{2}$

$\displaystyle \Rightarrow \frac{d^{2}z}{dx^{2}}=4>0$

$\displaystyle \text{So, } z \text{ is minimum at } x=\frac{15}{2}.$

$\displaystyle \text{Substituting } x=\frac{15}{2} \text{ in equation (1),}$

$\displaystyle y=\frac{15}{2}$

$\displaystyle \text{Thus, } z \text{ is minimum when } x=y=\frac{15}{2}.$

$\displaystyle \textbf{Question 2: }~\text{Divide }64\text{ into two parts such that the sum of the cubes of the } \\ \text{two parts is minimum.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Suppose } 64 \text{ is divided into two parts } x \text{ and } (64-x).$

$\displaystyle \text{Then, } z=x^{3}+(64-x)^{3}$

$\displaystyle \Rightarrow \frac{dz}{dx}=3x^{2}-3(64-x)^{2}$

$\displaystyle \text{For maximum or minimum values of } z, \text{ we must have } \frac{dz}{dx}=0$

$\displaystyle \Rightarrow 3x^{2}-3(64-x)^{2}=0$

$\displaystyle \Rightarrow x^{2}=(64-x)^{2}$

$\displaystyle \Rightarrow x=64-x$

$\displaystyle \Rightarrow x=32$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}z}{dx^{2}}=6x+6(64-x)$

$\displaystyle \Rightarrow \frac{d^{2}z}{dx^{2}}=384>0$

$\displaystyle \text{Thus, } z \text{ is minimum when } 64 \text{ is divided into two equal parts } 32 \text{ and } 32.$

$\displaystyle \textbf{Question 3: }~\text{How should we choose two numbers, each greater than or equal to }-2,\text{ whose sum is }1/2\text{ so that the sum of the first and the cube of the second is minimum?}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the two numbers be } x \text{ and } y.$

$\displaystyle x,y>-2 \text{ and } x+y=\frac{1}{2} \qquad \text{(1)}$

$\displaystyle \text{Now,}$

$\displaystyle z=x+y^{3}$

$\displaystyle \Rightarrow z=x+\left(\frac{1}{2}-x\right)^{3} \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow \frac{dz}{dx}=1-3\left(\frac{1}{2}-x\right)^{2}$

$\displaystyle \text{For maximum or minimum values of } z, \text{ we must have } \frac{dz}{dx}=0$

$\displaystyle \Rightarrow 1-3\left(\frac{1}{2}-x\right)^{2}=0$

$\displaystyle \Rightarrow \left(\frac{1}{2}-x\right)^{2}=\frac{1}{3}$

$\displaystyle \Rightarrow \frac{1}{2}-x=\pm\frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow x=\frac{1}{2}\mp\frac{1}{\sqrt{3}}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}z}{dx^{2}}=6\left(\frac{1}{2}-x\right)$

$\displaystyle \text{At } x=\frac{1}{2}+\frac{1}{\sqrt{3}}:$

$\displaystyle \frac{d^{2}z}{dx^{2}}=6\left(\frac{1}{2}-\frac{1}{2}-\frac{1}{\sqrt{3}}\right)=-\frac{6}{\sqrt{3}}<0$

$\displaystyle \text{So, } z \text{ is maximum when } x=\frac{1}{2}+\frac{1}{\sqrt{3}}.$

$\displaystyle \text{At } x=\frac{1}{2}-\frac{1}{\sqrt{3}}:$

$\displaystyle \frac{d^{2}z}{dx^{2}}=6\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{\sqrt{3}}\right)=\frac{6}{\sqrt{3}}>0$

$\displaystyle \text{So, } z \text{ is minimum when } x=\frac{1}{2}-\frac{1}{\sqrt{3}}.$

$\displaystyle \text{Since } x+y=\frac{1}{2},$

$\displaystyle \text{Substituting } x=\frac{1}{2}-\frac{1}{\sqrt{3}} \text{ in equation (1),}$

$\displaystyle y=\frac{1}{\sqrt{3}}$

$\displaystyle \text{So, the required two numbers are } \left(\frac{1}{2}-\frac{1}{\sqrt{3}}\right) \text{ and } \frac{1}{\sqrt{3}}.$

$\displaystyle \textbf{Question 4: }~\text{Divide }15\text{ into two parts such that the square of one multiplied with the } \\ \text{cube of the other is minimum.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the two numbers be } x \text{ and } y.$

$\displaystyle x+y=15 \qquad \text{(1)}$

$\displaystyle \text{Now,}$

$\displaystyle z=x^{2}y^{3}$

$\displaystyle \Rightarrow z=x^{2}(15-x)^{3} \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow \frac{dz}{dx}=2x(15-x)^{3}-3x^{2}(15-x)^{2}$

$\displaystyle \text{For maximum or minimum values of } z, \text{ we must have } \frac{dz}{dx}=0$

$\displaystyle \Rightarrow 2x(15-x)^{3}-3x^{2}(15-x)^{2}=0$

$\displaystyle \Rightarrow x(15-x)^{2}\left[2(15-x)-3x\right]=0$

$\displaystyle \Rightarrow 2(15-x)-3x=0$

$\displaystyle \Rightarrow 30-2x-3x=0$

$\displaystyle \Rightarrow 5x=30$

$\displaystyle \Rightarrow x=6$

$\displaystyle \Rightarrow y=15-6=9$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}z}{dx^{2}}=2(15-x)^{3}-12x(15-x)^{2}+6x^{2}(15-x)$

$\displaystyle \text{At } x=6:$

$\displaystyle \frac{d^{2}z}{dx^{2}}=2(9)^{3}-12(6)(9)^{2}+6(6)^{2}(9)=-2430<0$

$\displaystyle \text{Thus, } z \text{ is maximum when } x=6 \text{ and } y=9.$

$\displaystyle \text{So, the required two parts into which } 15 \text{ should be divided are } 6 \text{ and } 9.$

$\displaystyle \textbf{Question 5: }~\text{Of all the closed cylindrical cans (right circular) which enclose a given } \\ \text{volume of } 100\text{ cm}^3,\text{ which has the minimum surface area?}\;[\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } r \text{ and } h \text{ be the radius and height of the cylinder respectively.}$

$\displaystyle \text{Volume } (V) \text{ of the cylinder }=\pi r^{2}h$

$\displaystyle \Rightarrow 100=\pi r^{2}h$

$\displaystyle \Rightarrow h=\frac{100}{\pi r^{2}}$

$\displaystyle \text{Surface area } (S) \text{ of the cylinder }=2\pi r^{2}+2\pi rh$

$\displaystyle \Rightarrow S=2\pi r^{2}+2\pi r\left(\frac{100}{\pi r^{2}}\right)$

$\displaystyle \Rightarrow S=2\pi r^{2}+\frac{200}{r}$

$\displaystyle \Rightarrow \frac{dS}{dr}=4\pi r-\frac{200}{r^{2}}$

$\displaystyle \text{For maximum or minimum value of } S, \text{ we must have } \frac{dS}{dr}=0$

$\displaystyle \Rightarrow 4\pi r-\frac{200}{r^{2}}=0$

$\displaystyle \Rightarrow 4\pi r^{3}=200$

$\displaystyle \Rightarrow r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}S}{dr^{2}}=4\pi+\frac{400}{r^{3}}$

$\displaystyle \Rightarrow \frac{d^{2}S}{dr^{2}}>0 \text{ when } r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}$

$\displaystyle \text{Thus, the surface area is minimum when } r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}}.$

$\displaystyle \text{At } r=\left(\frac{50}{\pi}\right)^{\frac{1}{3}},$

$\displaystyle h=\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}} =2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}$

$\displaystyle \textbf{Question 6: }~\text{A beam is supported at the two ends and is uniformly loaded. The } \\ \text{bending moment } M\text{ at a distance }x\text{ from one end is given by:}$

$\displaystyle \textbf{(i)}~M=\dfrac{WL}{2}x-\dfrac{W}{2}x^2$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given : } M=\frac{WL}{2}x-\frac{W}{2}x^{2}$

$\displaystyle \Rightarrow \frac{dM}{dx}=\frac{WL}{2}-Wx$

$\displaystyle \text{For maximum or minimum value of } M, \text{ we must have } \frac{dM}{dx}=0$

$\displaystyle \Rightarrow \frac{WL}{2}-Wx=0$

$\displaystyle \Rightarrow \frac{WL}{2}=Wx$

$\displaystyle \Rightarrow x=\frac{L}{2}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}M}{dx^{2}}=-W<0$

$\displaystyle \text{So, } M \text{ is maximum at } x=\frac{L}{2}.$

$\displaystyle \textbf{(ii)}~M=\dfrac{Wx}{3}-\dfrac{W}{3}\dfrac{x^3}{L^2}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given : } M=\frac{Wx}{3}-\frac{Wx^{3}}{3L^{2}}$

$\displaystyle \Rightarrow \frac{dM}{dx}=\frac{W}{3}-\frac{Wx^{2}}{L^{2}}$

$\displaystyle \text{For maximum or minimum value of } M, \text{ we must have } \frac{dM}{dx}=0$

$\displaystyle \Rightarrow \frac{W}{3}-\frac{Wx^{2}}{L^{2}}=0$

$\displaystyle \Rightarrow \frac{W}{3}=\frac{Wx^{2}}{L^{2}}$

$\displaystyle \Rightarrow x^{2}=\frac{L^{2}}{3}$

$\displaystyle \Rightarrow x=\frac{L}{\sqrt{3}}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}M}{dx^{2}}=-\frac{2Wx}{L^{2}}$

$\displaystyle \Rightarrow \frac{d^{2}M}{dx^{2}}<0 \text{ at } x=\frac{L}{\sqrt{3}}$

$\displaystyle \text{So, } M \text{ is maximum at } x=\frac{L}{\sqrt{3}}.$

$\displaystyle \textbf{Question 7: }~\text{A wire of length }28\text{ m is to be cut into two pieces. One piece } \\ \text{is made into a square and the other into a circle. What should be the lengths } \\ \text{of the two pieces so that the combined area is minimum?}\;[\text{CBSE 2007, 2010}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Suppose the wire, which is to be made into a square and a circle, is cut into two pieces of lengths } \\ x \text{ m and } y \text{ m respectively.}$

$\displaystyle x+y=28 \qquad \text{(1)}$

$\displaystyle \text{Perimeter of square }=4(\text{side})=x$

$\displaystyle \Rightarrow \text{Side of square}=\frac{x}{4}$

$\displaystyle \Rightarrow \text{Area of square}=\left(\frac{x}{4}\right)^{2}=\frac{x^{2}}{16}$

$\displaystyle \text{Circumference of circle }=2\pi r=y$

$\displaystyle \Rightarrow r=\frac{y}{2\pi}$

$\displaystyle \Rightarrow \text{Area of circle}=\pi r^{2} =\pi\left(\frac{y}{2\pi}\right)^{2}=\frac{y^{2}}{4\pi}$

$\displaystyle \text{Now,}$

$\displaystyle z=\text{Area of square}+\text{Area of circle}$

$\displaystyle \Rightarrow z=\frac{x^{2}}{16}+\frac{y^{2}}{4\pi}$

$\displaystyle \Rightarrow z=\frac{x^{2}}{16}+\frac{(28-x)^{2}}{4\pi} \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow \frac{dz}{dx}=\frac{2x}{16}-\frac{2(28-x)}{4\pi}$

$\displaystyle \text{For maximum or minimum values of } z, \text{ we must have } \frac{dz}{dx}=0$

$\displaystyle \Rightarrow \frac{2x}{16}-\frac{2(28-x)}{4\pi}=0$

$\displaystyle \Rightarrow \frac{x}{4}=\frac{28-x}{\pi}$

$\displaystyle \Rightarrow \pi x+4x=112$

$\displaystyle \Rightarrow x(\pi+4)=112$

$\displaystyle \Rightarrow x=\frac{112}{\pi+4}$

$\displaystyle \Rightarrow y=28-\frac{112}{\pi+4}=\frac{28\pi}{\pi+4}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}z}{dx^{2}}=\frac{1}{8}+\frac{1}{2\pi}>0$

$\displaystyle \text{Thus, } z \text{ is minimum when } x=\frac{112}{\pi+4} \text{ and } y=\frac{28\pi}{\pi+4}.$

$\displaystyle \text{Hence, the lengths of the two pieces of wire are } \frac{112}{\pi+4}\text{ m and } \frac{28\pi}{\pi+4}\text{ m respectively.}$

$\displaystyle \textbf{Question 8: }~\text{A wire of length }20\text{ m is cut into two pieces. One is bent into a } \\ \text{square and the other into an equilateral triangle. Where should the wire be cut } \\ \text{so that the sum of the areas is minimum?}\;[\text{CBSE 2005}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Suppose the wire, which is to be made into a square and an equilateral triangle, is cut into two pieces of lengths } x \text{ and } y \text{ respectively.}$

$\displaystyle x+y=20 \qquad \text{(1)}$

$\displaystyle \text{Perimeter of square }=4(\text{side})=x$

$\displaystyle \Rightarrow \text{Side of square}=\frac{x}{4}$

$\displaystyle \Rightarrow \text{Area of square}=\left(\frac{x}{4}\right)^{2}=\frac{x^{2}}{16}$

$\displaystyle \text{Perimeter of equilateral triangle }=3(\text{side})=y$

$\displaystyle \Rightarrow \text{Side of triangle}=\frac{y}{3}$

$\displaystyle \Rightarrow \text{Area of triangle}=\frac{\sqrt{3}}{4}\left(\frac{y}{3}\right)^{2} =\frac{\sqrt{3}y^{2}}{36}$

$\displaystyle \text{Now,}$

$\displaystyle z=\text{Area of square}+\text{Area of triangle}$

$\displaystyle \Rightarrow z=\frac{x^{2}}{16}+\frac{\sqrt{3}y^{2}}{36}$

$\displaystyle \Rightarrow z=\frac{x^{2}}{16}+\frac{\sqrt{3}(20-x)^{2}}{36} \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow \frac{dz}{dx}=\frac{2x}{16}-\frac{2\sqrt{3}(20-x)}{36}$

$\displaystyle \text{For maximum or minimum values of } z, \text{ we must have } \frac{dz}{dx}=0$

$\displaystyle \Rightarrow \frac{2x}{16}-\frac{2\sqrt{3}(20-x)}{36}=0$

$\displaystyle \Rightarrow \frac{x}{8}=\frac{\sqrt{3}(20-x)}{18}$

$\displaystyle \Rightarrow 9x=4\sqrt{3}(20-x)$

$\displaystyle \Rightarrow 9x+4\sqrt{3}x=80\sqrt{3}$

$\displaystyle \Rightarrow x(9+4\sqrt{3})=80\sqrt{3}$

$\displaystyle \Rightarrow x=\frac{80\sqrt{3}}{9+4\sqrt{3}}$

$\displaystyle \Rightarrow y=20-\frac{80\sqrt{3}}{9+4\sqrt{3}} =\frac{180}{9+4\sqrt{3}}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}z}{dx^{2}}=\frac{1}{8}+\frac{\sqrt{3}}{18}>0$

$\displaystyle \text{Thus, } z \text{ is minimum when } x=\frac{80\sqrt{3}}{9+4\sqrt{3}} \text{ and } y=\frac{180}{9+4\sqrt{3}}.$

$\displaystyle \text{Hence, the wire of length } 20 \text{ cm should be cut into pieces of lengths } \frac{80\sqrt{3}}{9+4\sqrt{3}} \text{ cm and } \frac{180}{9+4\sqrt{3}} \text{ cm.}$

$\displaystyle \textbf{Question 9: }~\text{Given the sum of the perimeters of a square and a circle, show that } \\ \text{the sum of their areas is least when one side of the square equals } \\ \text{the diameter of the circle.}\;[\text{ CBSE 2005, 2011, 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the length of a side of the square and the radius of the circle be } x \text{ and } r \text{ respectively.}$

$\displaystyle \text{It is given that the sum of the perimeters of the square and the circle is constant.}$

$\displaystyle 4x+2\pi r=K \qquad \text{(where } K \text{ is a constant)}$

$\displaystyle \Rightarrow x=\frac{K-2\pi r}{4} \qquad \text{(1)}$

$\displaystyle \text{Now,}$

$\displaystyle A=x^{2}+\pi r^{2}$

$\displaystyle \Rightarrow A=\left(\frac{K-2\pi r}{4}\right)^{2}+\pi r^{2} =\frac{(K-2\pi r)^{2}}{16}+\pi r^{2} \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow \frac{dA}{dr} =\frac{2(K-2\pi r)(-2\pi)}{16}+2\pi r$

$\displaystyle \Rightarrow \frac{dA}{dr} =-\frac{\pi(K-2\pi r)}{4}+2\pi r$

$\displaystyle \text{For minimum value of } A, \text{ we must have } \frac{dA}{dr}=0$

$\displaystyle \Rightarrow -\frac{\pi(K-2\pi r)}{4}+2\pi r=0$

$\displaystyle \Rightarrow \frac{\pi(K-2\pi r)}{4}=2\pi r$

$\displaystyle \Rightarrow K-2\pi r=8r \qquad \text{(2)}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}A}{dr^{2}}=\frac{\pi^{2}}{2}+2\pi>0$

$\displaystyle \text{Hence, the sum of the areas } A \text{ is minimum when } K-2\pi r=8r.$

$\displaystyle \text{From equations (1) and (2),}$

$\displaystyle x=\frac{8r}{4}=2r$

$\displaystyle \text{Therefore, the side of the square is equal to the diameter of the circle.}$

$\displaystyle \textbf{Question 10: }~\text{Find the largest possible area of a right-angled triangle whose } \\ \text{hypotenuse is }5\text{ cm long.}\;[\text{CBSE 2000}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the base of the right-angled triangle be } x \text{ and its height be } y.$

$\displaystyle x^{2}+y^{2}=5^{2}$

$\displaystyle \Rightarrow y^{2}=25-x^{2}$

$\displaystyle \Rightarrow y=\sqrt{25-x^{2}}$

$\displaystyle \text{As the area of the triangle, } A=\frac{1}{2}xy$

$\displaystyle \Rightarrow A(x)=\frac{1}{2}x\sqrt{25-x^{2}}$

$\displaystyle \Rightarrow A(x)=\frac{x\sqrt{25-x^{2}}}{2}$

$\displaystyle \Rightarrow A'(x)=\frac{\sqrt{25-x^{2}}}{2}-\frac{x^{2}}{2\sqrt{25-x^{2}}}$

$\displaystyle \Rightarrow A'(x)=\frac{25-x^{2}-x^{2}}{2\sqrt{25-x^{2}}}$

$\displaystyle \Rightarrow A'(x)=\frac{25-2x^{2}}{2\sqrt{25-x^{2}}}$

$\displaystyle \text{For maxima or minima, we must have } A'(x)=0$

$\displaystyle \Rightarrow \frac{25-2x^{2}}{2\sqrt{25-x^{2}}}=0$

$\displaystyle \Rightarrow 25-2x^{2}=0$

$\displaystyle \Rightarrow 2x^{2}=25$

$\displaystyle \Rightarrow x=\frac{5}{\sqrt{2}}$

$\displaystyle \Rightarrow y=\sqrt{25-\frac{25}{2}}=\sqrt{\frac{25}{2}}=\frac{5}{\sqrt{2}}$

$\displaystyle \text{Now,}$

$\displaystyle A''(x)=\frac{-75+2x^{3}}{(25-x^{2})\sqrt{25-x^{2}}}$

$\displaystyle A''\!\left(\frac{5}{\sqrt{2}}\right)<0$

$\displaystyle \text{So, } x=\frac{5}{\sqrt{2}} \text{ is the point of maximum area.}$

$\displaystyle \text{Hence, the largest possible area of the triangle is}$

$\displaystyle A_{\max}=\frac{1}{2}\cdot\frac{5}{\sqrt{2}}\cdot\frac{5}{\sqrt{2}}=\frac{25}{4}\text{ square units.}$

$\displaystyle \textbf{Question 11: }~\text{Two sides of a triangle have lengths }a\text{ and }b\text{ and the angle } \\ \text{between them is }\theta.\text{ What value of }\theta\text{ will maximize the area? } \\ \text{Find the maximum area also.}\;[\text{CBSE 2002 C}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{As the area of the triangle, } A=\frac{1}{2}ab\sin\theta$

$\displaystyle \Rightarrow A(\theta)=\frac{1}{2}ab\sin\theta$

$\displaystyle \Rightarrow A'(\theta)=\frac{1}{2}ab\cos\theta$

$\displaystyle \text{For maxima or minima, } A'(\theta)=0$

$\displaystyle \Rightarrow \frac{1}{2}ab\cos\theta=0$

$\displaystyle \Rightarrow \cos\theta=0$

$\displaystyle \Rightarrow \theta=\frac{\pi}{2}$

$\displaystyle \text{Also,}$

$\displaystyle A''(\theta)=-\frac{1}{2}ab\sin\theta$

$\displaystyle A''\!\left(\frac{\pi}{2}\right)=-\frac{1}{2}ab\sin\!\left(\frac{\pi}{2}\right) =-\frac{1}{2}ab<0$

$\displaystyle \text{Hence, } \theta=\frac{\pi}{2} \text{ is the point of maximum area.}$

$\displaystyle \text{Now,}$

$\displaystyle A_{\max}=\frac{1}{2}ab\sin\!\left(\frac{\pi}{2}\right)=\frac{ab}{2}$

$\displaystyle \textbf{Question 12: }~\text{A square piece of tin of side }18\text{ cm is to be made into a box without } \\ \text{top by cutting a square from each corner and folding up the flaps. What should } \\ \text{be the side of the square cut off so that the volume of the box is maximum? Also, } \\ \text{find this maximum volume.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the side of the square to be cut off be } x \text{ cm.}$

$\displaystyle \text{Then, the length and breadth of the box will be } (18-2x) \text{ cm each,}$

$\displaystyle \text{and the height of the box will be } x \text{ cm.}$

$\displaystyle \text{Volume of the box, } V(x)=x(18-2x)^{2}$

$\displaystyle \Rightarrow V'(x)=(18-2x)^{2}-4x(18-2x)$

$\displaystyle \Rightarrow V'(x)=(18-2x)(18-2x-4x)$

$\displaystyle \Rightarrow V'(x)=(18-2x)(18-6x)$

$\displaystyle \Rightarrow V'(x)=12(9-x)(3-x)$

$\displaystyle \text{For maximum or minimum values of } V, \text{ we must have } V'(x)=0$

$\displaystyle \Rightarrow x=9 \text{ or } x=3$

$\displaystyle \text{If } x=9, \text{ then the length and breadth become zero.}$

$\displaystyle \Rightarrow x\neq 9$

$\displaystyle \Rightarrow x=3$

$\displaystyle \text{Now,}$

$\displaystyle V''(x)=-24(6-x)$

$\displaystyle V''(3)=-24(6-3)=-72<0$

$\displaystyle \text{So, } x=3 \text{ is the point of maximum volume.}$

$\displaystyle \text{The maximum volume is given by}$

$\displaystyle V(3)=3(18-6)^{2}=3\times144=432\text{ cm}^{3}$

$\displaystyle \text{Hence, if a square of side } 3 \text{ cm is removed from each corner of the square tin,}$

$\displaystyle \text{the volume of the box obtained is maximum and is equal to } 432\text{ cm}^{3}.$

$\displaystyle \textbf{Question 13: }~\text{A rectangular sheet of tin }45\text{ cm by }24\text{ cm is to be made into a } \\ \text{box without top by cutting off squares from each corner and folding up the flaps. What } \\ \text{should be the side of the square to be cut off so that the volume of the box is } \\ \text{maximum possible?}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Suppose a square of side measuring } x \text{ cm is cut off from each corner.}$

$\displaystyle \text{Then, the length, breadth and height of the box will be } (45-2x),\ (24-2x) \text{ and } x \text{ respectively.}$

$\displaystyle \text{Volume of the box, } V=(45-2x)(24-2x)x$

$\displaystyle \Rightarrow \frac{dV}{dx} =(45-2x)(24-2x)-2x(45-2x)-2x(24-2x)$

$\displaystyle \text{For maximum or minimum values of } V, \text{ we must have } \frac{dV}{dx}=0$

$\displaystyle \Rightarrow (45-2x)(24-2x)-2x(45-2x)-2x(24-2x)=0$

$\displaystyle \Rightarrow 4x^{2}-138x+1080-90x+4x^{2}-48x+4x^{2}=0$

$\displaystyle \Rightarrow 12x^{2}-276x+1080=0$

$\displaystyle \Rightarrow x^{2}-23x+90=0$

$\displaystyle \Rightarrow (x-18)(x-5)=0$

$\displaystyle \Rightarrow x=18 \text{ or } x=5$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}V}{dx^{2}}=24x-276$

$\displaystyle \frac{d^{2}V}{dx^{2}}\Big|_{x=5}=120-276=-156<0$

$\displaystyle \frac{d^{2}V}{dx^{2}}\Big|_{x=18}=432-276=156>0$

$\displaystyle \text{Thus, the volume of the box is maximum when } x=5\text{ cm.}$

$\displaystyle \text{Hence, the side of the square to be cut off measures } 5\text{ cm.}$

$\displaystyle \textbf{Question 14: }~\text{A tank with rectangular base and rectangular sides, open at the } \\ \text{top, is to be constructed so that its depth is }2\text{ m and volume is }8\text{ m}^3.\text{ If cost of } \\ \text{tank is Rs }70\text{ per square metre for the base and Rs }45\text{ per square } \\ \text{metre for sides, find the cost of least expensive tank.}\;[\text{ CBSE 2009}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } l,\ b \text{ and } h \text{ be the length, breadth and height of the tank respectively.}$

$\displaystyle \text{Height } h=2\text{ m}$

$\displaystyle \text{Volume of the tank }=8\text{ m}^3$

$\displaystyle \Rightarrow l\times b\times h=8$

$\displaystyle \Rightarrow l\times b\times 2=8$

$\displaystyle \Rightarrow lb=4$

$\displaystyle \Rightarrow b=\frac{4}{l}$

$\displaystyle \text{Area of the base }=lb=4\text{ m}^2$

$\displaystyle \text{Area of the four walls }=2h(l+b)$

$\displaystyle \Rightarrow A=2(2)\left(l+\frac{4}{l}\right)$

$\displaystyle \Rightarrow A=4\left(l+\frac{4}{l}\right)$

$\displaystyle \Rightarrow \frac{dA}{dl}=4\left(1-\frac{4}{l^{2}}\right)$

$\displaystyle \text{For maximum or minimum values of } A,\ \text{we must have } \frac{dA}{dl}=0$

$\displaystyle \Rightarrow 4\left(1-\frac{4}{l^{2}}\right)=0$

$\displaystyle \Rightarrow l^{2}=4$

$\displaystyle \Rightarrow l=\pm 2$

$\displaystyle \text{Since length cannot be negative, } l=2\text{ m}$

$\displaystyle \Rightarrow b=\frac{4}{2}=2\text{ m}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}A}{dl^{2}}=\frac{32}{l^{3}}$

$\displaystyle \Rightarrow \frac{d^{2}A}{dl^{2}}\Big|_{l=2}=\frac{32}{8}=4>0$

$\displaystyle \text{Thus, the area is minimum when } l=2\text{ m}$

$\displaystyle \text{Hence, } l=b=h=2\text{ m}$

$\displaystyle \text{Cost of building the base }= \text{Rs }70\times(lb)=\text{Rs }70\times4=\text{Rs }280$

$\displaystyle \text{Cost of building the walls }=\text{Rs }8\times\text{(area of walls)}$

$\displaystyle =\text{Rs }8\times\left[2h(l+b)\right] =\text{Rs }8\times\left[2(2)(2+2)\right]=\text{Rs }8\times16=\text{Rs }720$

$\displaystyle \text{Total cost }=\text{Rs }280+\text{Rs }720=\text{Rs }1000$

$\displaystyle \text{Hence, the total cost of the tank will be Rs }1000.$

$\displaystyle \textbf{Question 15: }~\text{A window in the form of a rectangle is surmounted by a semicircular } \\ \text{opening. The total perimeter is }10\text{ m. Find the dimensions of the rectangular part to } \\ \text{admit maximum light.}\;[\text{ CBSE 2000, 2002, 2011, 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the dimensions of the rectangular part be } x \text{ and } y.$

$\displaystyle \text{Radius of the semi-circle}=\frac{x}{2}$

$\displaystyle \text{Total perimeter }=10$

$\displaystyle \Rightarrow (x+2y)+\pi\left(\frac{x}{2}\right)=10$

$\displaystyle \Rightarrow 2y=10-x-\frac{\pi x}{2}$

$\displaystyle \Rightarrow y=\frac{1}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right] \qquad \text{(1)}$

$\displaystyle \text{Now,}$

$\displaystyle \text{Area } A=\text{Area of rectangle}+\text{Area of semi-circle}$

$\displaystyle \Rightarrow A=xy+\frac{1}{2}\pi\left(\frac{x}{2}\right)^{2}$

$\displaystyle \Rightarrow A=xy+\frac{\pi x^{2}}{8}$

$\displaystyle \Rightarrow A=\frac{\pi x^{2}}{8} +\frac{x}{2}\left[10-x\left(1+\frac{\pi}{2}\right)\right] \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow A=\frac{\pi x^{2}}{8}+\frac{10x}{2} -\frac{x^{2}}{2}\left(1+\frac{\pi}{2}\right)$

$\displaystyle \Rightarrow \frac{dA}{dx} =\frac{\pi x}{4}+5-x\left(1+\frac{\pi}{2}\right)$

$\displaystyle \text{For maximum or minimum values of } A,\ \text{we must have } \frac{dA}{dx}=0$

$\displaystyle \Rightarrow \frac{\pi x}{4}+5-x\left(1+\frac{\pi}{2}\right)=0$

$\displaystyle \Rightarrow x\left(\frac{\pi}{4}-1-\frac{\pi}{2}\right)=-5$

$\displaystyle \Rightarrow x\left(-\frac{\pi+4}{4}\right)=-5$

$\displaystyle \Rightarrow x=\frac{20}{\pi+4}$

$\displaystyle \text{Substituting the value of } x \text{ in equation (1),}$

$\displaystyle y=\frac{1}{2}\left[10-\frac{20}{\pi+4} \left(1+\frac{\pi}{2}\right)\right]$

$\displaystyle \Rightarrow y=\frac{10}{\pi+4}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}A}{dx^{2}} =\frac{\pi}{4}-\left(1+\frac{\pi}{2}\right) =-\frac{\pi+4}{4}<0$

$\displaystyle \text{Thus, the area is maximum when } x=\frac{20}{\pi+4} \text{ and } y=\frac{10}{\pi+4}.$

$\displaystyle \text{So, the required dimensions are:}$

$\displaystyle \text{Length }=\frac{20}{\pi+4}\text{ m}$

$\displaystyle \text{Breadth }=\frac{10}{\pi+4}\text{ m}$

$\displaystyle \textbf{Question 16: }~\text{A large window has the shape of a rectangle surmounted by an } \\ \text{equilateral triangle. If the perimeter is }12\text{ m, find the dimensions of the rectangle } \\ \text{that produce the largest area of the window.}\;[\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the dimensions of the rectangular part of the window be } x \text{ and } y.$

$\displaystyle \text{Perimeter of the window }=x+y+x+x+y=12$

$\displaystyle \Rightarrow 3x+2y=12$

$\displaystyle \Rightarrow y=\frac{12-3x}{2} \qquad \text{(1)}$

$\displaystyle \text{Area of the window }=\text{Area of rectangle}+\text{Area of semicircle}$

$\displaystyle \Rightarrow A=xy+\frac{\sqrt{3}}{4}x^{2}$

$\displaystyle \Rightarrow A=x\left(\frac{12-3x}{2}\right)+\frac{\sqrt{3}}{4}x^{2}$

$\displaystyle \Rightarrow A=6x-\frac{3x^{2}}{2}+\frac{\sqrt{3}}{4}x^{2}$

$\displaystyle \Rightarrow \frac{dA}{dx} =6-3x+\frac{\sqrt{3}}{2}x$

$\displaystyle \Rightarrow \frac{dA}{dx} =6-x\left(3-\frac{\sqrt{3}}{2}\right)$

$\displaystyle \text{For maximum or minimum values of } A, \text{ we must have } \frac{dA}{dx}=0$

$\displaystyle \Rightarrow 6=x\left(3-\frac{\sqrt{3}}{2}\right)$

$\displaystyle \Rightarrow x=\frac{12}{6-\sqrt{3}}$

$\displaystyle \text{Substituting the value of } x \text{ in equation (1),}$

$\displaystyle y=\frac{12-3\left(\frac{12}{6-\sqrt{3}}\right)}{2}$

$\displaystyle \Rightarrow y=\frac{18-6\sqrt{3}}{6-\sqrt{3}}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}A}{dx^{2}}=-3+\frac{\sqrt{3}}{2}<0$

$\displaystyle \text{Thus, the area of the window is maximum when }$

$\displaystyle x=\frac{12}{6-\sqrt{3}} \text{ and } y=\frac{18-6\sqrt{3}}{6-\sqrt{3}}.$

$\displaystyle \textbf{Question 17: }~\text{Show that the height of the cylinder of maximum volume that can be } \\ \text{inscribed in a sphere of radius }R\text{ is }\dfrac{2R}{\sqrt{3}}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the height and radius of the base of the cylinder be } h \text{ and } r \text{ respectively.}$

$\displaystyle \frac{h^{2}}{4}+r^{2}=R^{2}$

$\displaystyle \Rightarrow h=2\sqrt{R^{2}-r^{2}} \qquad \text{(1)}$

$\displaystyle \text{Volume of the cylinder } V=\pi r^{2}h$

$\displaystyle \text{Squaring both sides, we get}$

$\displaystyle V^{2}=\pi^{2}r^{4}h^{2}$

$\displaystyle \Rightarrow V^{2}=4\pi^{2}r^{4}(R^{2}-r^{2}) \qquad \text{[From equation (1)]}$

$\displaystyle \text{Let } Z=4\pi^{2}(r^{4}R^{2}-r^{6})$

$\displaystyle \Rightarrow \frac{dZ}{dr}=4\pi^{2}(4r^{3}R^{2}-6r^{5})$

$\displaystyle \text{For maximum or minimum values of } Z, \text{ we must have } \frac{dZ}{dr}=0$

$\displaystyle \Rightarrow 4r^{3}R^{2}-6r^{5}=0$

$\displaystyle \Rightarrow 2r^{3}(2R^{2}-3r^{2})=0$

$\displaystyle \Rightarrow 2R^{2}-3r^{2}=0$

$\displaystyle \Rightarrow r^{2}=\frac{2R^{2}}{3}$

$\displaystyle \Rightarrow r=\frac{R\sqrt{6}}{3}$

$\displaystyle \text{Substituting the value of } r \text{ in equation (1),}$

$\displaystyle h=2\sqrt{R^{2}-\frac{2R^{2}}{3}}$

$\displaystyle \Rightarrow h=2\sqrt{\frac{R^{2}}{3}}$

$\displaystyle \Rightarrow h=\frac{2R}{\sqrt{3}}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}Z}{dr^{2}}=4\pi^{2}(12r^{2}R^{2}-30r^{4})$

$\displaystyle \Rightarrow \frac{d^{2}Z}{dr^{2}}\Bigg|_{r^{2}=\frac{2R^{2}}{3}} =4\pi^{2}\left(8R^{4}-\frac{40R^{4}}{3}\right) =-\frac{16\pi^{2}R^{4}}{3}<0$

$\displaystyle \text{Thus, the volume of the cylinder is maximum when } h=\frac{2R}{\sqrt{3}}.$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 18: }~\text{A rectangle is inscribed in a semicircle of radius }r\text{ with one side } \\ \text{on the diameter. Find the dimensions so that its area is maximum. Find also } \\ \text{the area.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the dimensions of the rectangle be } x \text{ and } y.$

$\displaystyle \frac{x^{2}}{4}+y^{2}=r^{2}$

$\displaystyle \Rightarrow x^{2}+4y^{2}=4r^{2}$

$\displaystyle \Rightarrow x^{2}=4(r^{2}-y^{2}) \qquad \text{(1)}$

$\displaystyle \text{Area of the rectangle }=xy$

$\displaystyle \Rightarrow A=xy$

$\displaystyle \text{Squaring both sides, we get}$

$\displaystyle A^{2}=x^{2}y^{2}$

$\displaystyle \Rightarrow Z=4y^{2}(r^{2}-y^{2}) \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow \frac{dZ}{dy}=8yr^{2}-16y^{3}$

$\displaystyle \text{For maximum or minimum values of } Z, \text{ we must have } \frac{dZ}{dy}=0$

$\displaystyle \Rightarrow 8yr^{2}-16y^{3}=0$

$\displaystyle \Rightarrow 8y(r^{2}-2y^{2})=0$

$\displaystyle \Rightarrow r^{2}=2y^{2}$

$\displaystyle \Rightarrow y=\frac{r}{\sqrt{2}}$

$\displaystyle \text{Substituting the value of } y \text{ in equation (1),}$

$\displaystyle x^{2}=4\left(r^{2}-\frac{r^{2}}{2}\right)$

$\displaystyle \Rightarrow x^{2}=2r^{2}$

$\displaystyle \Rightarrow x=r\sqrt{2}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}Z}{dy^{2}}=8r^{2}-48y^{2}$

$\displaystyle \Rightarrow \frac{d^{2}Z}{dy^{2}}\Bigg|_{y=\frac{r}{\sqrt{2}}} =8r^{2}-48\left(\frac{r^{2}}{2}\right) =-16r^{2}<0$

$\displaystyle \text{So, the area is maximum when } x=r\sqrt{2} \text{ and } y=\frac{r}{\sqrt{2}}.$

$\displaystyle \text{Hence, the maximum area is}$

$\displaystyle A_{\max}=xy=r\sqrt{2}\times\frac{r}{\sqrt{2}}=r^{2}.$

$\displaystyle \textbf{Question 19: }~\text{Prove that a conical tent of given capacity will require the least amount } \\ \text{of canvas when the height is }\sqrt{2}\text{ times the radius of the base.}\;[\text{ CBSE 2007, 2011, 2013}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the surface area of the conical tent be } S=\pi r\sqrt{r^{2}+h^{2}}.$

$\displaystyle \text{Let the volume of the conical tent be } V=\frac{1}{3}\pi r^{2}h.$

$\displaystyle \Rightarrow h=\frac{3V}{\pi r^{2}}.$

$\displaystyle \Rightarrow S=\pi r\sqrt{r^{2}+\left(\frac{3V}{\pi r^{2}}\right)^{2}}.$

$\displaystyle \Rightarrow S=\pi r\sqrt{r^{2}+\frac{9V^{2}}{\pi^{2}r^{4}}} =\frac{1}{r}\sqrt{\pi^{2}r^{6}+9V^{2}}.$

$\displaystyle \text{Now, differentiating with respect to } r,$

$\displaystyle \frac{dS}{dr} =\frac{d}{dr}\left[\frac{1}{r}\sqrt{\pi^{2}r^{6}+9V^{2}}\right]$

$\displaystyle \Rightarrow \frac{dS}{dr} =\frac{6\pi^{2}r^{5}}{2\sqrt{\pi^{2}r^{6}+9V^{2}}} -\frac{\sqrt{\pi^{2}r^{6}+9V^{2}}}{r^{2}}.$

$\displaystyle \text{For minimum value of } S,\ \text{we must have } \frac{dS}{dr}=0.$

$\displaystyle \Rightarrow \frac{3\pi^{2}r^{5}}{\sqrt{\pi^{2}r^{6}+9V^{2}}} =\frac{\sqrt{\pi^{2}r^{6}+9V^{2}}}{r^{2}}.$

$\displaystyle \Rightarrow 3\pi^{2}r^{7}=\pi^{2}r^{6}+9V^{2}.$

$\displaystyle \Rightarrow 2\pi^{2}r^{6}=9V^{2}.$

$\displaystyle \text{Substituting the value of } V=\frac{1}{3}\pi r^{2}h,$

$\displaystyle 2\pi^{2}r^{6} =9\left(\frac{1}{3}\pi r^{2}h\right)^{2}.$

$\displaystyle \Rightarrow 2\pi^{2}r^{6}=\pi^{2}r^{4}h^{2}.$

$\displaystyle \Rightarrow 2r^{2}=h^{2}.$

$\displaystyle \Rightarrow h=\sqrt{2}\,r.$

$\displaystyle \textbf{Question 20: }~\text{Show that the cone of the greatest volume which can be inscribed in a } \\ \text{of given sphere has an altitude equal to }2/3\text{ of the diameter of the sphere.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } h,\ r \text{ and } R \text{ be the height, radius of the base of the cone and radius of the sphere respectively.}$

$\displaystyle h=R+\sqrt{R^{2}-r^{2}}$

$\displaystyle \Rightarrow (h-R)^{2}=R^{2}-r^{2}$

$\displaystyle \Rightarrow h^{2}+R^{2}-2hR=R^{2}-r^{2}$

$\displaystyle \Rightarrow r^{2}=2hR-h^{2} \qquad \text{(1)}$

$\displaystyle \text{Volume of the cone } V=\frac{1}{3}\pi r^{2}h$

$\displaystyle \Rightarrow V=\frac{1}{3}\pi h(2hR-h^{2}) \qquad \text{[From equation (1)]}$

$\displaystyle \Rightarrow V=\frac{\pi}{3}(2h^{2}R-h^{3})$

$\displaystyle \Rightarrow \frac{dV}{dh}=\frac{\pi}{3}(4hR-3h^{2})$

$\displaystyle \text{For maximum or minimum values of } V,\ \text{we must have } \frac{dV}{dh}=0$

$\displaystyle \Rightarrow 4hR-3h^{2}=0$

$\displaystyle \Rightarrow h(4R-3h)=0$

$\displaystyle \Rightarrow h=\frac{4R}{3}$

$\displaystyle \text{Now,}$

$\displaystyle \frac{d^{2}V}{dh^{2}}=\frac{\pi}{3}(4R-6h)$

$\displaystyle \Rightarrow \frac{d^{2}V}{dh^{2}}\Big|_{h=\frac{4R}{3}} =\frac{\pi}{3}\left(4R-6\cdot\frac{4R}{3}\right) =-\frac{4\pi R}{3}<0$

$\displaystyle \text{So, the volume of the cone is maximum when } h=\frac{4R}{3}.$

$\displaystyle \Rightarrow h=\frac{2}{3}\times(\text{diameter of the sphere}).$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 21: }~\text{Prove that the semi-vertical angle of the right circular cone of given volume } \\ \text{and least curved surface is }\cot^{-1}(\sqrt{2}).\;[\text{CBSE 2014}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let:}$

$\displaystyle \text{Radius of the base }=r,$

$\displaystyle \text{Height }=h,$

$\displaystyle \text{Slant height }=l,$

$\displaystyle \text{Volume }=V,$

$\displaystyle \text{Curved surface area }=C.$

$\displaystyle \text{As, volume of the cone } V=\frac{1}{3}\pi r^{2}h$

$\displaystyle \Rightarrow h=\frac{3V}{\pi r^{2}}$

$\displaystyle \text{Also, slant height } l=\sqrt{h^{2}+r^{2}}$

$\displaystyle \Rightarrow l=\sqrt{\left(\frac{3V}{\pi r^{2}}\right)^{2}+r^{2}}$

$\displaystyle \Rightarrow l=\sqrt{\frac{9V^{2}}{\pi^{2}r^{4}}+r^{2}}$

$\displaystyle \Rightarrow l=\sqrt{\frac{9V^{2}+\pi^{2}r^{6}}{\pi^{2}r^{4}}}$

$\displaystyle \Rightarrow l=\frac{\sqrt{9V^{2}+\pi^{2}r^{6}}}{\pi r^{2}}$

$\displaystyle \text{Now, curved surface area } C=\pi rl$

$\displaystyle \Rightarrow C(r)=\pi r\cdot\frac{\sqrt{9V^{2}+\pi^{2}r^{6}}}{\pi r^{2}}$

$\displaystyle \Rightarrow C(r)=\frac{\sqrt{9V^{2}+\pi^{2}r^{6}}}{r}$

$\displaystyle \Rightarrow C'(r)=\frac{r\cdot\frac{6\pi^{2}r^{5}}{2\sqrt{9V^{2}+\pi^{2}r^{6}}}-\sqrt{9V^{2}+\pi^{2}r^{6}}}{r^{2}}$

$\displaystyle \Rightarrow C'(r)=\frac{3\pi^{2}r^{6}-9V^{2}}{r^{2}\sqrt{9V^{2}+\pi^{2}r^{6}}}$

$\displaystyle \text{For maximum or minimum value of } C,\ \text{we must have } C'(r)=0$

$\displaystyle \Rightarrow 3\pi^{2}r^{6}-9V^{2}=0$

$\displaystyle \Rightarrow \pi^{2}r^{6}=3V^{2}$

$\displaystyle \Rightarrow V=\frac{\pi r^{3}}{\sqrt{3}}$

$\displaystyle \text{Substituting in } h=\frac{3V}{\pi r^{2}},$

$\displaystyle h=\frac{3}{\pi r^{2}}\cdot\frac{\pi r^{3}}{\sqrt{3}}=r\sqrt{3}$

$\displaystyle \Rightarrow \frac{h}{r}=\sqrt{3}$

$\displaystyle \Rightarrow \cot\theta=\sqrt{3}$

$\displaystyle \Rightarrow \theta=\cot^{-1}(\sqrt{3})$

$\displaystyle \text{Thus, the curved surface area of the cone is minimum when } h=r\sqrt{3}.$

$\displaystyle \textbf{Question 22: }~\text{An isosceles triangle of vertical angle }2\theta\text{ is inscribed in a circle of radius }a. \\ \text{ Show that the area of the triangle is maximum when }\theta=\dfrac{\pi}{6}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } ABC \text{ be an isosceles triangle inscribed in a circle of radius } a \text{ such that } AB=AC.$

$\displaystyle \text{Let } \angle BAC=2\theta.$

$\displaystyle AD=AO+OD=a+a\cos 2\theta=a(1+\cos 2\theta)$

$\displaystyle BC=2BD=2a\sin 2\theta$

$\displaystyle \text{Area of the triangle } A=\frac12 \times BC \times AD$

$\displaystyle A(\theta)=\frac12 \cdot 2a\sin 2\theta \cdot a(1+\cos 2\theta)$

$\displaystyle A(\theta)=a^{2}\sin 2\theta(1+\cos 2\theta)$

$\displaystyle A(\theta)=a^{2}\sin 2\theta+a^{2}\sin 2\theta\cos 2\theta$

$\displaystyle A(\theta)=a^{2}\sin 2\theta+\frac{a^{2}}{2}\sin 4\theta$

$\displaystyle A'(\theta)=2a^{2}\cos 2\theta+2a^{2}\cos 4\theta$

$\displaystyle A'(\theta)=2a^{2}(\cos 2\theta+\cos 4\theta)$

$\displaystyle \text{For maxima or minima, } A'(\theta)=0$

$\displaystyle \Rightarrow \cos 2\theta+\cos 4\theta=0$

$\displaystyle \Rightarrow \cos 2\theta=-\cos 4\theta$

$\displaystyle \Rightarrow \cos 2\theta=\cos(\pi-4\theta)$

$\displaystyle \Rightarrow 2\theta=\pi-4\theta$

$\displaystyle \Rightarrow 6\theta=\pi$

$\displaystyle \Rightarrow \theta=\frac{\pi}{6}$

$\displaystyle \text{Now, } A''(\theta)=2a^{2}(-\sin 2\theta-2\sin 4\theta)$

$\displaystyle A''\!\left(\frac{\pi}{6}\right)<0$

$\displaystyle \text{Hence, the area of the triangle is maximum when } \theta=\frac{\pi}{6}.$

$\displaystyle \textbf{Question 23: }~\text{Prove that the least perimeter of an isosceles triangle in which a circle } \\ \text{of radius }r\text{ can be inscribed is }6\sqrt{3}\,r.\;[\text{CBSE 2016}]$
$\displaystyle \text{Answer:}$

$\displaystyle \textbf{To prove:}~\text{The least perimeter of an isosceles triangle in which a circle of radius } r \text{ can be inscribed is } 6\sqrt{3}\,r.$

$\displaystyle \text{Let } ABC \text{ be an isosceles triangle with } AB=AC=x \text{ and } BC=y.$

$\displaystyle \text{Let a circle of radius } r \text{ be inscribed in } \triangle ABC \text{ with centre } O.$

$\displaystyle \text{Since } O \text{ is the incentre, it divides the median in the ratio } 2:1.$

$\displaystyle \Rightarrow AO=2r \text{ and } OF=r.$

$\displaystyle \text{Using Pythagoras theorem in } \triangle ABF:$

$\displaystyle AF^{2}+BF^{2}=AB^{2}$

$\displaystyle \Rightarrow (3r)^{2}+\left(\frac{y}{2}\right)^{2}=x^{2}\quad\text{(1)}$

$\displaystyle \text{Again, from } \triangle ADO,$

$\displaystyle (2r)^{2}=r^{2}+AD^{2}$

$\displaystyle \Rightarrow 3r^{2}=AD^{2}$

$\displaystyle \Rightarrow AD=\sqrt{3}\,r$

$\displaystyle \text{Now, } DB=BF \text{ and } EC=FC \text{ (tangents from an external point are equal).}$

$\displaystyle \Rightarrow AD+DB=x$

$\displaystyle \Rightarrow \sqrt{3}\,r+\frac{y}{2}=x$

$\displaystyle \Rightarrow \frac{y}{2}=x-\sqrt{3}\,r\quad\text{(2)}$

$\displaystyle \text{Substituting (2) in (1):}$

$\displaystyle (3r)^{2}+(x-\sqrt{3}\,r)^{2}=x^{2}$

$\displaystyle \Rightarrow 9r^{2}+x^{2}-2\sqrt{3}rx+3r^{2}=x^{2}$

$\displaystyle \Rightarrow 12r^{2}=2\sqrt{3}rx$

$\displaystyle \Rightarrow 6r=\sqrt{3}x$

$\displaystyle \Rightarrow x=\frac{6r}{\sqrt{3}}$

$\displaystyle \text{Now, from (2):}$

$\displaystyle \frac{y}{2}=\frac{6r}{\sqrt{3}}-\sqrt{3}r$

$\displaystyle \Rightarrow \frac{y}{2}=\frac{6r-3r}{\sqrt{3}}$

$\displaystyle \Rightarrow \frac{y}{2}=\sqrt{3}r$

$\displaystyle \Rightarrow y=2\sqrt{3}r$

$\displaystyle \text{Perimeter } =2x+y$

$\displaystyle =2\left(\frac{6r}{\sqrt{3}}\right)+2\sqrt{3}r$

$\displaystyle =\frac{12r}{\sqrt{3}}+2\sqrt{3}r$

$\displaystyle =\frac{12r+6r}{\sqrt{3}}$

$\displaystyle =\frac{18r}{\sqrt{3}}$

$\displaystyle =6\sqrt{3}\,r$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 24: }~\text{Find the dimensions of the rectangle of perimeter }36\text{ cm which will } \\ \text{sweep out a volume as large as possible when revolved about one of its sides.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } l, b \text{ and } V \text{ be the length, breadth and volume of the rectangle respectively.}$

$\displaystyle 2(l+b)=36$

$\displaystyle \Rightarrow l=18-b\quad\text{(1)}$

$\displaystyle \text{Volume of the solid formed when the rectangle is revolved about its breadth is that of a cylinder.}$

$\displaystyle V=\pi l^{2}b$

$\displaystyle \Rightarrow V=\pi(18-b)^{2}b\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow V=\pi(324b-36b^{2}+b^{3})$

$\displaystyle \frac{dV}{db}=\pi(324-72b+3b^{2})$

$\displaystyle \text{For maximum or minimum values of } V,\ \frac{dV}{db}=0$

$\displaystyle \Rightarrow 324-72b+3b^{2}=0$

$\displaystyle \Rightarrow b^{2}-24b+108=0$

$\displaystyle \Rightarrow b^{2}-6b-18b+108=0$

$\displaystyle \Rightarrow (b-6)(b-18)=0$

$\displaystyle \Rightarrow b=6 \text{ or } b=18$

$\displaystyle \text{Now, } \frac{d^{2}V}{db^{2}}=\pi(6b-72)$

$\displaystyle \text{At } b=6:$

$\displaystyle \frac{d^{2}V}{db^{2}}=\pi(36-72)=-36\pi<0$

$\displaystyle \text{Hence, } V \text{ is maximum at } b=6.$

$\displaystyle \text{At } b=18:$

$\displaystyle \frac{d^{2}V}{db^{2}}=\pi(108-72)=36\pi>0$

$\displaystyle \text{Hence, } V \text{ is minimum at } b=18.$

$\displaystyle \text{Substituting } b=6 \text{ in eq. (1), we get}$

$\displaystyle l=18-6=12$

$\displaystyle \text{Therefore, the volume is maximum when } l=12\text{ cm and } b=6\text{ cm.}$

$\displaystyle \textbf{Question 25: }~\text{Show that the height of the cone of maximum volume that } \\ \text{can be inscribed in a sphere of radius }12\text{ cm is }16\text{ cm.}\;[\text{CBSE 2005}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the height, radius of base and volume of the cone be } h, r \text{ and } V \text{ respectively.}$

$\displaystyle h=R+\sqrt{R^{2}-r^{2}}$

$\displaystyle \Rightarrow h-R=\sqrt{R^{2}-r^{2}}$

$\displaystyle \text{Squaring both sides, we get}$

$\displaystyle h^{2}+R^{2}-2hR=R^{2}-r^{2}$

$\displaystyle \Rightarrow r^{2}=2hR-h^{2}\quad\text{(1)}$

$\displaystyle \text{Now, volume of the cone } V=\frac13\pi r^{2}h$

$\displaystyle \Rightarrow V=\frac{\pi}{3}(2hR-h^{2})h\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow V=\frac{\pi}{3}(2h^{2}R-h^{3})$

$\displaystyle \frac{dV}{dh}=\frac{\pi}{3}(4hR-3h^{2})$

$\displaystyle \text{For maximum or minimum values of } V,\ \frac{dV}{dh}=0$

$\displaystyle \Rightarrow 4hR-3h^{2}=0$

$\displaystyle \Rightarrow h(4R-3h)=0$

$\displaystyle \Rightarrow h=\frac{4R}{3}$

$\displaystyle \text{Now, } \frac{d^{2}V}{dh^{2}}=\frac{\pi}{3}(4R-6h)$

$\displaystyle \text{At } h=\frac{4R}{3}:$

$\displaystyle \frac{d^{2}V}{dh^{2}}=\frac{\pi}{3}(4R-8R)=-\frac{4\pi R}{3}<0$

$\displaystyle \text{Hence, the volume of the cone is maximum when } h=\frac{4R}{3}.$

$\displaystyle \text{If } R=12\text{ cm, then } h=\frac{4\times12}{3}=16\text{ cm.}$

$\displaystyle \textbf{Question 26: }~\text{A closed cylinder has volume }2156\text{ cm}^3.\text{ What } \\ \text{will be the radius of its base so that its total surface area is minimum?}\;[\text{CBSE 2000 C}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the height, radius of the base and surface area of the cylinder be } h, r \text{ and } S \text{ respectively.}$

$\displaystyle \text{Volume of the cylinder }= \pi r^{2}h$

$\displaystyle \Rightarrow 2156=\pi r^{2}h$

$\displaystyle \Rightarrow 2156=\frac{22}{7}r^{2}h$

$\displaystyle \Rightarrow h=\frac{2156\times7}{22r^{2}}$

$\displaystyle \Rightarrow h=\frac{686}{r^{2}}\quad\text{(1)}$

$\displaystyle \text{Surface area of the cylinder } S=2\pi rh+2\pi r^{2}$

$\displaystyle \Rightarrow S=2\pi r\left(\frac{686}{r^{2}}\right)+2\pi r^{2}\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow S=\frac{1372\pi}{r}+2\pi r^{2}$

$\displaystyle \Rightarrow S=\frac{4312}{r}+\frac{44r^{2}}{7}$

$\displaystyle \frac{dS}{dr}=-\frac{4312}{r^{2}}+\frac{88r}{7}$

$\displaystyle \text{For maximum or minimum values of } S,\ \frac{dS}{dr}=0$

$\displaystyle \Rightarrow -\frac{4312}{r^{2}}+\frac{88r}{7}=0$

$\displaystyle \Rightarrow \frac{4312}{r^{2}}=\frac{88r}{7}$

$\displaystyle \Rightarrow r^{3}=\frac{4312\times7}{88}$

$\displaystyle \Rightarrow r^{3}=343$

$\displaystyle \Rightarrow r=7\text{ cm}$

$\displaystyle \text{Now, } \frac{d^{2}S}{dr^{2}}=\frac{8624}{r^{3}}+\frac{88}{7}$

$\displaystyle \text{At } r=7:$

$\displaystyle \frac{d^{2}S}{dr^{2}}=\frac{8624}{343}+\frac{88}{7}$

$\displaystyle \Rightarrow \frac{d^{2}S}{dr^{2}}=\frac{176}{7}>0$

$\displaystyle \text{Hence, the surface area is minimum when } r=7\text{ cm.}$

$\displaystyle \textbf{Question 27: }~\text{Show that the maximum volume of the cylinder which can be } \\ \text{inscribed in a sphere of radius }5\sqrt{3}\text{ cm is }500\pi\text{ cm}^3.\;[\text{CBSE 2004}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the height, radius of base and volume of a cylinder be } h, r \text{ and } V \text{ respectively.}$

$\displaystyle \frac{h^{2}}{4}+r^{2}=R^{2}$

$\displaystyle \Rightarrow h^{2}=4(R^{2}-r^{2})$

$\displaystyle \Rightarrow r^{2}=R^{2}-\frac{h^{2}}{4}\quad\text{(1)}$

$\displaystyle \text{Now, volume of the cylinder } V=\pi r^{2}h$

$\displaystyle \Rightarrow V=\pi h\left(R^{2}-\frac{h^{2}}{4}\right)\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow V=\pi\left(R^{2}h-\frac{h^{3}}{4}\right)$

$\displaystyle \frac{dV}{dh}=\pi\left(R^{2}-\frac{3h^{2}}{4}\right)$

$\displaystyle \text{For maximum or minimum values of } V,\ \frac{dV}{dh}=0$

$\displaystyle \Rightarrow R^{2}-\frac{3h^{2}}{4}=0$

$\displaystyle \Rightarrow R^{2}=\frac{3h^{2}}{4}$

$\displaystyle \Rightarrow h=\frac{2R}{\sqrt{3}}$

$\displaystyle \text{Now, } \frac{d^{2}V}{dh^{2}}=-\frac{3\pi h}{2}$

$\displaystyle \text{At } h=\frac{2R}{\sqrt{3}}:$

$\displaystyle \frac{d^{2}V}{dh^{2}}=-\frac{3\pi}{2}\cdot\frac{2R}{\sqrt{3}}=-\frac{3\pi R}{\sqrt{3}}<0$

$\displaystyle \text{Hence, the volume of the cylinder is maximum when } h=\frac{2R}{\sqrt{3}}.$

$\displaystyle \text{Maximum volume }= \pi h\left(R^{2}-\frac{h^{2}}{4}\right)$

$\displaystyle =\pi\cdot\frac{2R}{\sqrt{3}}\left(R^{2}-\frac{4R^{2}}{12}\right)$

$\displaystyle =\pi\cdot\frac{2R}{\sqrt{3}}\cdot\frac{2R^{2}}{3}$

$\displaystyle =\frac{4\pi R^{3}}{3\sqrt{3}}$

$\displaystyle \text{If } R=5\sqrt{3}, \text{ then}$

$\displaystyle \text{Maximum volume }=\frac{4\pi(5\sqrt{3})^{3}}{3\sqrt{3}}=500\pi\text{ cm}^{3}$

$\displaystyle \textbf{Question 28: }~\text{Show that among all positive numbers }x\text{ and }y\text{ with }\\ x^2+y^2=r^2, \text{ the sum }x+y\text{ is largest when }x=y=\dfrac{r}{\sqrt{2}}.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Here,}$

$\displaystyle x^{2}+y^{2}=r^{2}$

$\displaystyle \Rightarrow y=\sqrt{r^{2}-x^{2}}\quad\text{(1)}$

$\displaystyle \text{Now, let } Z=x+y$

$\displaystyle \Rightarrow Z=x+\sqrt{r^{2}-x^{2}}\quad\text{[From eq. (1)]}$

$\displaystyle \frac{dZ}{dx}=1+\frac{-2x}{2\sqrt{r^{2}-x^{2}}}$

$\displaystyle \text{For maximum or minimum values of } Z,\ \frac{dZ}{dx}=0$

$\displaystyle \Rightarrow 1-\frac{x}{\sqrt{r^{2}-x^{2}}}=0$

$\displaystyle \Rightarrow x=\sqrt{r^{2}-x^{2}}$

$\displaystyle \text{Squaring both sides, we get}$

$\displaystyle x^{2}=r^{2}-x^{2}$

$\displaystyle \Rightarrow 2x^{2}=r^{2}$

$\displaystyle \Rightarrow x=\frac{r}{\sqrt{2}}$

$\displaystyle \text{Substituting the value of } x \text{ in eq. (1), we get}$

$\displaystyle y=\sqrt{r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}}$

$\displaystyle \Rightarrow y=\frac{r}{\sqrt{2}}$

$\displaystyle \text{Now, } \frac{d^{2}Z}{dx^{2}}=-\frac{r^{2}}{(r^{2}-x^{2})^{3/2}}$

$\displaystyle \text{At } x=\frac{r}{\sqrt{2}}:$

$\displaystyle \frac{d^{2}Z}{dx^{2}}=-\frac{r^{2}}{\left(\frac{r^{2}}{2}\right)^{3/2}}=-\frac{2\sqrt{2}}{r}<0$

$\displaystyle \text{Hence, } Z=x+y \text{ is maximum when } x=y=\frac{r}{\sqrt{2}}.$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 29: }~\text{Determine the points on the curve }x^2=4y\text{ which are nearest to } \\ \text{the point }(0,5).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the point }(x,y)\text{ on the curve }x^{2}=4y\text{ be nearest to the point }(0,5).$

$\displaystyle x^{2}=4y$

$\displaystyle \Rightarrow y=\frac{x^{2}}{4}\quad\text{(1)}$

$\displaystyle \text{Using the distance formula,}$

$\displaystyle d^{2}=x^{2}+(y-5)^{2}$

$\displaystyle \text{Let } Z=d^{2}$

$\displaystyle \Rightarrow Z=x^{2}+(y-5)^{2}$

$\displaystyle \Rightarrow Z=x^{2}+\left(\frac{x^{2}}{4}-5\right)^{2}\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow Z=x^{2}+\frac{x^{4}}{16}-\frac{5x^{2}}{2}+25$

$\displaystyle \Rightarrow Z=\frac{x^{4}}{16}-\frac{3x^{2}}{2}+25$

$\displaystyle \frac{dZ}{dx}=\frac{4x^{3}}{16}-3x$

$\displaystyle \text{For maximum or minimum values of } Z,\ \frac{dZ}{dx}=0$

$\displaystyle \Rightarrow \frac{x^{3}}{4}-3x=0$

$\displaystyle \Rightarrow x\left(\frac{x^{2}}{4}-3\right)=0$

$\displaystyle \Rightarrow x=0 \text{ or } x^{2}=12$

$\displaystyle \Rightarrow x=\pm 2\sqrt{3}$

$\displaystyle \text{Substituting the value of } x \text{ in eq. (1), we get}$

$\displaystyle y=\frac{(2\sqrt{3})^{2}}{4}=3$

$\displaystyle \text{Now, } \frac{d^{2}Z}{dx^{2}}=\frac{3x^{2}}{4}-3$

$\displaystyle \text{At } x=\pm 2\sqrt{3}:$

$\displaystyle \frac{d^{2}Z}{dx^{2}}=\frac{3(12)}{4}-3=6>0$

$\displaystyle \text{Hence, the point on the curve nearest to }(0,5)\text{ is }(\pm 2\sqrt{3},\,3).$

$\displaystyle \textbf{Question 30: }~\text{Find the point on the curve }y^2=4x\text{ which is nearest } \\ \text{to the point }(2,-8).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the point }(x,y)\text{ on the curve be nearest to the point }(2,-8).$

$\displaystyle y^{2}=4x$

$\displaystyle \Rightarrow x=\frac{y^{2}}{4}\quad\text{(1)}$

$\displaystyle \text{Using the distance formula,}$

$\displaystyle d^{2}=(x-2)^{2}+(y+8)^{2}$

$\displaystyle \text{Let } Z=d^{2}$

$\displaystyle \Rightarrow Z=(x-2)^{2}+(y+8)^{2}$

$\displaystyle \Rightarrow Z=\left(\frac{y^{2}}{4}-2\right)^{2}+(y+8)^{2}\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow Z=\frac{y^{4}}{16}-y^{2}+4+y^{2}+64+16y$

$\displaystyle \Rightarrow Z=\frac{y^{4}}{16}+16y+68$

$\displaystyle \frac{dZ}{dy}=\frac{4y^{3}}{16}+16$

$\displaystyle \text{For maximum or minimum values of } Z,\ \frac{dZ}{dy}=0$

$\displaystyle \Rightarrow \frac{y^{3}}{4}+16=0$

$\displaystyle \Rightarrow y^{3}=-64$

$\displaystyle \Rightarrow y=-4$

$\displaystyle \text{Substituting the value of } y \text{ in eq. (1), we get}$

$\displaystyle x=\frac{(-4)^{2}}{4}=4$

$\displaystyle \text{Now, } \frac{d^{2}Z}{dy^{2}}=\frac{12y^{2}}{16}$

$\displaystyle \text{At } y=-4:$

$\displaystyle \frac{d^{2}Z}{dy^{2}}=\frac{12(16)}{16}=12>0$

$\displaystyle \text{Hence, the point on the curve nearest to }(2,-8)\text{ is }(4,-4).$

$\displaystyle \textbf{Question 31: }~\text{Find the point on the curve }x^2=8y\text{ which is nearest to } \\ \text{the point }(2,4).\;[\text{CBSE 2007}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the point }(x,y)\text{ on the curve be nearest to the point }(2,4).$

$\displaystyle x^{2}=8y$

$\displaystyle \Rightarrow y=\frac{x^{2}}{8}\quad\text{(1)}$

$\displaystyle \text{Using the distance formula,}$

$\displaystyle d^{2}=(x-2)^{2}+(y-4)^{2}$

$\displaystyle \text{Let } Z=d^{2}$

$\displaystyle \Rightarrow Z=(x-2)^{2}+(y-4)^{2}$

$\displaystyle \Rightarrow Z=(x-2)^{2}+\left(\frac{x^{2}}{8}-4\right)^{2}\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow Z=x^{2}-4x+4+\frac{x^{4}}{64}-x^{2}+16$

$\displaystyle \Rightarrow Z=\frac{x^{4}}{64}-4x+20$

$\displaystyle \frac{dZ}{dx}=\frac{4x^{3}}{64}-4$

$\displaystyle \text{For maximum or minimum values of } Z,\ \frac{dZ}{dx}=0$

$\displaystyle \Rightarrow \frac{x^{3}}{16}-4=0$

$\displaystyle \Rightarrow x^{3}=64$

$\displaystyle \Rightarrow x=4$

$\displaystyle \text{Substituting the value of } x \text{ in eq. (1), we get}$

$\displaystyle y=\frac{4^{2}}{8}=2$

$\displaystyle \text{Now, } \frac{d^{2}Z}{dx^{2}}=\frac{12x^{2}}{64}$

$\displaystyle \text{At } x=4:$

$\displaystyle \frac{d^{2}Z}{dx^{2}}=\frac{12(16)}{64}=3>0$

$\displaystyle \text{Hence, the point on the curve nearest to }(2,4)\text{ is }(4,2).$

$\displaystyle \textbf{Question 32: }~\text{Find the point on the parabola }x^2=2y\text{ which is closest } \\ \text{to the point }(0,5).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the required point be }(x,y).$

$\displaystyle x^{2}=2y$

$\displaystyle \Rightarrow y=\frac{x^{2}}{2}\quad\text{(1)}$

$\displaystyle \text{The distance between the points }(x,y)\text{ and }(0,5)\text{ is given by}$

$\displaystyle d^{2}=x^{2}+(y-5)^{2}$

$\displaystyle \text{Let } Z=d^{2}$

$\displaystyle \Rightarrow Z=x^{2}+(y-5)^{2}$

$\displaystyle \Rightarrow Z=x^{2}+\left(\frac{x^{2}}{2}-5\right)^{2}\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow Z=x^{2}+\frac{x^{4}}{4}-5x^{2}+25$

$\displaystyle \Rightarrow Z=\frac{x^{4}}{4}-4x^{2}+25$

$\displaystyle \frac{dZ}{dx}=x^{3}-8x$

$\displaystyle \text{For maximum or minimum values of } Z,\ \frac{dZ}{dx}=0$

$\displaystyle \Rightarrow x^{3}-8x=0$

$\displaystyle \Rightarrow x(x^{2}-8)=0$

$\displaystyle \Rightarrow x=\pm2\sqrt{2}$

$\displaystyle \text{Substituting the value of } x \text{ in eq. (1), we get}$

$\displaystyle y=\frac{(2\sqrt{2})^{2}}{2}=4$

$\displaystyle \text{Now, } \frac{d^{2}Z}{dx^{2}}=3x^{2}-8$

$\displaystyle \text{At } x=\pm2\sqrt{2}:$

$\displaystyle \frac{d^{2}Z}{dx^{2}}=24-8=16>0$

$\displaystyle \text{Hence, the point on the curve nearest to }(0,5)\text{ is }(\pm2\sqrt{2},\,4).$

$\displaystyle \textbf{Question 33: }~\text{Find the coordinates of a point on the parabola } \\ y=x^2+7x+2\text{ which is closest to the straight line }y=3x-3.\;[\text{CBSE 2015}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the coordinates of the point on the parabola be }(x,y).$

$\displaystyle y=x^{2}+7x+2\quad\text{(1)}$

$\displaystyle \text{The distance of the point }(x,\ x^{2}+7x+2)\text{ from the line }y=3x-3\text{ is }S.$

$\displaystyle S=\frac{\left|-3x+(x^{2}+7x+2)+3\right|}{\sqrt{10}}$

$\displaystyle S=\frac{|x^{2}+4x+5|}{\sqrt{10}}$

$\displaystyle \text{To minimise }S,\text{ minimise }S^{2}.$

$\displaystyle \Rightarrow S^{2}=\frac{(x^{2}+4x+5)^{2}}{10}$

$\displaystyle \frac{d(S^{2})}{dx}=\frac{2(x^{2}+4x+5)(2x+4)}{10}$

$\displaystyle \text{For minimum value, } \frac{d(S^{2})}{dx}=0$

$\displaystyle \Rightarrow (x^{2}+4x+5)(2x+4)=0$

$\displaystyle \Rightarrow 2x+4=0$

$\displaystyle \Rightarrow x=-2$

$\displaystyle \text{Substituting }x=-2\text{ in eq. (1),}$

$\displaystyle y=(-2)^{2}+7(-2)+2=4-14+2=-8$

$\displaystyle \text{Now, } \frac{d^{2}(S^{2})}{dx^{2}}=\frac{2(2x+4)^{2}+4(x^{2}+4x+5)}{10}$

$\displaystyle \text{At }x=-2,\ \frac{d^{2}(S^{2})}{dx^{2}}>0$

$\displaystyle \text{Hence, the point on the parabola nearest to the line is }(-2,-8).$

$\displaystyle \textbf{Question 34: }~\text{Find the point on the curve }y^2=2x\text{ which is at a minimum } \\ \text{distance from the point }(1,4).\;[\text{CBSE 2011}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Suppose a point }(x,y)\text{ on the curve }y^{2}=2x\text{ is nearest to the point }(1,4).$

$\displaystyle y^{2}=2x$

$\displaystyle \Rightarrow x=\frac{y^{2}}{2}\quad\text{(1)}$

$\displaystyle \text{Using the distance formula,}$

$\displaystyle d^{2}=(x-1)^{2}+(y-4)^{2}$

$\displaystyle \text{Let } Z=d^{2}$

$\displaystyle \Rightarrow Z=(x-1)^{2}+(y-4)^{2}$

$\displaystyle \Rightarrow Z=\left(\frac{y^{2}}{2}-1\right)^{2}+(y-4)^{2}\quad\text{[From eq. (1)]}$

$\displaystyle \Rightarrow Z=\frac{y^{4}}{4}-y^{2}+1+y^{2}-8y+16$

$\displaystyle \Rightarrow Z=\frac{y^{4}}{4}-8y+17$

$\displaystyle \frac{dZ}{dy}=y^{3}-8$

$\displaystyle \text{For maximum or minimum values of } Z,\ \frac{dZ}{dy}=0$

$\displaystyle \Rightarrow y^{3}-8=0$

$\displaystyle \Rightarrow y^{3}=8$

$\displaystyle \Rightarrow y=2$

$\displaystyle \text{Substituting the value of } y \text{ in eq. (1), we get}$

$\displaystyle x=\frac{2^{2}}{2}=2$

$\displaystyle \text{Now, } \frac{d^{2}Z}{dy^{2}}=3y^{2}$

$\displaystyle \text{At } y=2:$

$\displaystyle \frac{d^{2}Z}{dy^{2}}=12>0$

$\displaystyle \text{Hence, the point on the curve nearest to }(1,4)\text{ is }(2,2).$

$\displaystyle \textbf{Question 35: }~\text{Find the maximum slope of the curve } y=-x^3+3x^2+2x-27.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } y=-x^{3}+3x^{2}+2x-27\quad\text{(1)}$

$\displaystyle \text{Slope }=\frac{dy}{dx}=-3x^{2}+6x+2$

$\displaystyle \text{Let } M=-3x^{2}+6x+2$

$\displaystyle \Rightarrow \frac{dM}{dx}=-6x+6$

$\displaystyle \text{For maximum or minimum values of } M,\ \frac{dM}{dx}=0$

$\displaystyle \Rightarrow -6x+6=0$

$\displaystyle \Rightarrow 6x=6$

$\displaystyle \Rightarrow x=1$

$\displaystyle \text{Substituting } x=1 \text{ in eq. (1), we get}$

$\displaystyle y=-(1)^{3}+3(1)^{2}+2(1)-27$

$\displaystyle \Rightarrow y=-1+3+2-27=-23$

$\displaystyle \text{Now, } \frac{d^{2}M}{dx^{2}}=-6<0$

$\displaystyle \text{Hence, the slope is maximum at } x=1.$

$\displaystyle \text{At the point }(1,-23):$

$\displaystyle \text{Maximum slope }=M(1)=-3(1)^{2}+6(1)+2$

$\displaystyle \Rightarrow \text{Maximum slope }=5$

$\displaystyle \textbf{Question 36: }~\text{The total cost of producing }x\text{ radio sets per day is } \\ \text{Rs }\left(\dfrac{x^2}{4}+35x+25\right)\text{ and the price per set is Rs }\left(50-\dfrac{x}{2}\right). \\ \text{ Find the daily output to maximize the total profit.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Profit }= \text{S.P.} - \text{C.P.}$

$\displaystyle \Rightarrow P=x\left(50-\frac{x}{2}\right)-\left(\frac{x^{2}}{4}+35x+25\right)$

$\displaystyle \Rightarrow P=50x-\frac{x^{2}}{2}-\frac{x^{2}}{4}-35x-25$

$\displaystyle \Rightarrow P=15x-\frac{3x^{2}}{4}-25$

$\displaystyle \frac{dP}{dx}=15-\frac{3x}{2}$

$\displaystyle \text{For maximum or minimum values of } P,\ \frac{dP}{dx}=0$

$\displaystyle \Rightarrow 15-\frac{3x}{2}=0$

$\displaystyle \Rightarrow 15=\frac{3x}{2}$

$\displaystyle \Rightarrow x=10$

$\displaystyle \text{Now, } \frac{d^{2}P}{dx^{2}}=-\frac{3}{2}<0$

$\displaystyle \text{Hence, the profit is maximum when the daily output is }10\text{ items.}$

$\displaystyle \textbf{Question 37: }~\text{A manufacturer can sell }x\text{ items at a price of } \\ \text{Rs }\left(5-\dfrac{x}{100}\right)\text{ each. The cost price is Rs }\left(\dfrac{x}{5}+500\right). \\ \text{ Find the number of items he should sell to earn maximum profit.}\;[\text{CBSE 2009}]$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Profit }= \text{S.P.} - \text{C.P.}$

$\displaystyle \Rightarrow P=x\left(5-\frac{x}{100}\right)-\left(500+\frac{x}{5}\right)$

$\displaystyle \Rightarrow P=5x-\frac{x^{2}}{100}-500-\frac{x}{5}$

$\displaystyle \Rightarrow P=\frac{24x}{5}-\frac{x^{2}}{100}-500$

$\displaystyle \frac{dP}{dx}=\frac{24}{5}-\frac{x}{50}$

$\displaystyle \text{For maximum or minimum values of } P,\ \frac{dP}{dx}=0$

$\displaystyle \Rightarrow \frac{24}{5}-\frac{x}{50}=0$

$\displaystyle \Rightarrow \frac{x}{50}=\frac{24}{5}$

$\displaystyle \Rightarrow x=\frac{24\times50}{5}$

$\displaystyle \Rightarrow x=240$

$\displaystyle \text{Now, } \frac{d^{2}P}{dx^{2}}=-\frac{1}{50}<0$

$\displaystyle \text{Hence, the profit is maximum when }240\text{ items are sold.}$

$\displaystyle \textbf{Question 38: }~\text{An open tank is to be constructed with a square base and } \\ \text{vertical sides so as to contain a given quantity of water. Show that the expenses of lining } \\ \text{with lead will be least if the depth is made half of the width.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } l, h, V \text{ and } S \text{ be the length, height, volume and surface area of the tank respectively.}$

$\displaystyle \text{Since the volume } V \text{ is constant,}$

$\displaystyle l^{2}h = V$

$\displaystyle \Rightarrow h = \frac{V}{l^{2}} \qquad (1)$

$\displaystyle \text{Surface area, } S = l^{2} + 4lh$

$\displaystyle \Rightarrow S = l^{2} + \frac{4V}{l} \qquad \text{[From eq. (1)]}$

$\displaystyle \frac{dS}{dl} = 2l - \frac{4V}{l^{2}}$

$\displaystyle \text{For } S \text{ to be maximum or minimum, } \frac{dS}{dl} = 0$

$\displaystyle \Rightarrow 2l - \frac{4V}{l^{2}} = 0$

$\displaystyle \Rightarrow 2l^{3} - 4V = 0$

$\displaystyle \Rightarrow l^{3} = 2V$

$\displaystyle \text{Now, } \frac{d^{2}S}{dl^{2}} = 2 + \frac{8V}{l^{3}}$

$\displaystyle \Rightarrow \frac{d^{2}S}{dl^{2}} = 2 + \frac{8V}{2V} = 6 > 0$

$\displaystyle \text{Hence, the surface area is minimum.}$

$\displaystyle h = \frac{V}{l^{2}}$

$\displaystyle \text{Substituting } V = \frac{l^{3}}{2} \text{ in eq. (1),}$

$\displaystyle h = \frac{l^{3}}{2l^{2}} = \frac{l}{2}$

$\displaystyle \text{Hence proved.}$

$\displaystyle \textbf{Question 39: }~\text{A box of constant volume is to be twice as long as it is wide. The } \\ \text{material on the top and four sides costs three times as much per square metre as } \\ \text{that in the bottom. What are the most economic dimensions?}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } l, b \text{ and } h \text{ be the length, breadth and height of the box respectively.}$

$\displaystyle \text{Volume of the box } = c$

$\displaystyle \text{Given: } l = 2b \qquad (1)$

$\displaystyle \Rightarrow c = lbh$

$\displaystyle \Rightarrow c = 2b^{2}h$

$\displaystyle \Rightarrow h = \frac{c}{2b^{2}} \qquad (2)$

$\displaystyle \text{Let the cost of the material required for the bottom be } K \text{ per } m^{2}.$

$\displaystyle \text{Cost of the material required for the four walls and the top be } 3K \text{ per } m^{2}.$

$\displaystyle \text{Total cost } T = K(lb) + 3K(2lh + 2bh + lb)$

$\displaystyle \Rightarrow T = 2Kb^{2} + 3K\!\left(\frac{4bc}{2b^{2}} + \frac{2bc}{2b^{2}} + 2b^{2}\right) \qquad \text{[From eqs. (1) and (2)]}$

$\displaystyle \Rightarrow T = 2Kb^{2} + 3K\!\left(\frac{3c}{b} + 2b^{2}\right)$

$\displaystyle \frac{dT}{db} = 4Kb + 3K\!\left(-\frac{3c}{b^{2}} + 4b\right)$

$\displaystyle \text{For } T \text{ to be maximum or minimum, } \frac{dT}{db} = 0$

$\displaystyle \Rightarrow 4Kb + 3K\!\left(-\frac{3c}{b^{2}} + 4b\right) = 0$

$\displaystyle \Rightarrow 4b = 3\!\left(\frac{3c}{b^{2}} - 4b\right)$

$\displaystyle \Rightarrow 4b = \frac{9c}{b^{2}} - 12b$

$\displaystyle \Rightarrow 16b = \frac{9c}{b^{2}}$

$\displaystyle \Rightarrow 16b^{3} = 9c$

$\displaystyle \Rightarrow b = \left(\frac{9c}{16}\right)^{\frac{1}{3}}$

$\displaystyle \text{Now, } \frac{d^{2}T}{db^{2}} = 4K + 3K\!\left(\frac{6c}{b^{3}} + 4\right)$

$\displaystyle \Rightarrow \frac{d^{2}T}{db^{2}} = 4K + 3K\!\left(\frac{6c}{\frac{9c}{16}} + 4\right)$

$\displaystyle \Rightarrow \frac{d^{2}T}{db^{2}} = 4K + 3K\!\left(\frac{32}{3} + 4\right) = 48K > 0$

$\displaystyle \therefore \text{Cost is minimum when } b = \left(\frac{9c}{16}\right)^{\frac{1}{3}}.$

$\displaystyle \text{Substituting } b = \left(\frac{9c}{16}\right)^{\frac{1}{3}} \text{ in eqs. (1) and (2),}$

$\displaystyle l = 2\left(\frac{9c}{16}\right)^{\frac{1}{3}}$

$\displaystyle h = \frac{c}{2b^{2}} = \frac{c}{2\left(\frac{9c}{16}\right)^{\frac{2}{3}}} = \left(\frac{32c}{81}\right)^{\frac{1}{3}}$

$\displaystyle \text{Thus, the most economical dimensions of the box are }$

$\displaystyle l = 2\left(\frac{9c}{16}\right)^{\frac{1}{3}}, \quad b = \left(\frac{9c}{16}\right)^{\frac{1}{3}}, \quad h = \left(\frac{32c}{81}\right)^{\frac{1}{3}}.$

$\displaystyle \textbf{Question 40: }~\text{The sum of the surface areas of a sphere and a cube is given. Show } \\ \text{that when the sum of their volumes is least, the diameter of the sphere is equal } \\ \text{to the edge of the cube.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } r \text{ be the radius of the sphere, } x \text{ be the side of the cube and } S \text{ be the sum of the surface areas of both.}$

$\displaystyle S = 4\pi r^{2} + 6x^{2}$

$\displaystyle \Rightarrow x = \left(\frac{S - 4\pi r^{2}}{6}\right)^{\frac{1}{2}} \qquad (1)$

$\displaystyle \text{Sum of volumes, } V = \frac{4}{3}\pi r^{3} + x^{3}$

$\displaystyle \Rightarrow V = \frac{4}{3}\pi r^{3} + \left(\frac{S - 4\pi r^{2}}{6}\right)^{\frac{3}{2}} \qquad \text{[From eq. (1)]}$

$\displaystyle \frac{dV}{dr} = 4\pi r^{2} - 2\pi r\left(\frac{S - 4\pi r^{2}}{6}\right)^{\frac{1}{2}}$

$\displaystyle \text{For maximum or minimum value of } V, \text{ we must have } \frac{dV}{dr} = 0 \qquad (2)$

$\displaystyle \Rightarrow 4\pi r^{2} - 2\pi r\left(\frac{S - 4\pi r^{2}}{6}\right)^{\frac{1}{2}} = 0 \qquad \text{[From eq. (2)]}$

$\displaystyle \Rightarrow 4\pi r^{2} = 2\pi r\left(\frac{S - 4\pi r^{2}}{6}\right)^{\frac{1}{2}}$

$\displaystyle \Rightarrow 2r = \left(\frac{S - 4\pi r^{2}}{6}\right)^{\frac{1}{2}}$

$\displaystyle \Rightarrow x = 2r \qquad \text{[From eq. (1)]}$

$\displaystyle \text{Now, } \frac{d^{2}V}{dr^{2}} = 8\pi r - 2\pi\left(\frac{S - 4\pi r^{2}}{6}\right)^{\frac{1}{2}} + \frac{4}{3}\pi^{2} r^{2}\left(\frac{6}{S - 4\pi r^{2}}\right)^{\frac{1}{2}}$

$\displaystyle \Rightarrow \frac{d^{2}V}{dr^{2}} = 8\pi r - 4\pi r + \frac{2}{3}\pi^{2} r$

$\displaystyle \Rightarrow \frac{d^{2}V}{dr^{2}} = 4\pi r + \frac{2}{3}\pi^{2} r > 0$

$\displaystyle \text{Hence, the volume is minimum when } x = 2r.$

$\displaystyle \textbf{Question 41: }~\text{A given quantity of metal is to be cast into a half cylinder with a } \\ \text{rectangular base and semicircular ends. Show that in order that the total surface area } \\ \text{may be minimum, the ratio of the length of the cylinder to the diameter of its } \\ \text{semicircular ends is }\pi:(\pi+2).$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Volume, } V = \frac{1}{2}\pi l\left(\frac{D}{2}\right)^{2}$

$\displaystyle \Rightarrow V = \frac{\pi D^{2}l}{8}$

$\displaystyle \Rightarrow l = \frac{8V}{\pi D^{2}} \qquad (1)$

$\displaystyle \text{Total surface area, } S = \frac{\pi D^{2}}{4} + lD + \frac{\pi Dl}{2}$

$\displaystyle \Rightarrow S = \frac{\pi D^{2}}{4} + \frac{8V}{\pi D} + \frac{4V}{D} \qquad \text{[From eq. (1)]}$

$\displaystyle \frac{dS}{dD} = \frac{\pi D}{2} - \frac{8V}{\pi D^{2}} - \frac{4V}{D^{2}}$

$\displaystyle \text{For maximum or minimum value of } S, \text{ we must have } \frac{dS}{dD} = 0$

$\displaystyle \Rightarrow \frac{\pi D}{2} = \frac{8V}{\pi D^{2}} + \frac{4V}{D^{2}}$

$\displaystyle \Rightarrow \frac{\pi D^{3}}{2} = 4V\!\left(\frac{2}{\pi} + 1\right)$

$\displaystyle \Rightarrow D^{3} = \frac{8V}{\pi}\!\left(\frac{2}{\pi} + 1\right)$

$\displaystyle \text{Now, } \frac{d^{2}S}{dD^{2}} = \frac{\pi}{2} + \frac{16V}{\pi D^{3}} + \frac{8V}{D^{3}}$

$\displaystyle \Rightarrow \frac{d^{2}S}{dD^{2}} = \frac{\pi}{2} + \pi > 0$

$\displaystyle \text{Hence, the surface area is minimum.}$

$\displaystyle l = \frac{8V}{\pi D^{2}}$

$\displaystyle \Rightarrow l = \frac{8}{\pi D^{2}}\!\left[\frac{D^{3}}{8}\cdot\frac{\pi^{2}}{\pi + 2}\right]$

$\displaystyle \Rightarrow l = D\!\left(\frac{\pi}{\pi + 2}\right)$

$\displaystyle \Rightarrow \frac{l}{D} = \frac{\pi}{\pi + 2}$

$\displaystyle \text{Hence, proved.}$

$\displaystyle \textbf{Question 42: }~\text{The strength of a beam varies as the product of its breadth and } \\ \text{square of its depth. Find the dimensions of the strongest beam which can be cut from } \\ \text{a circular log of radius }a.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the breadth, height and strength of the beam be } b, h \text{ and } S \text{ respectively.}$

$\displaystyle a^{2} = \frac{h^{2} + b^{2}}{4}$

$\displaystyle \Rightarrow 4a^{2} = h^{2} + b^{2}$

$\displaystyle \Rightarrow h^{2} = 4a^{2} - b^{2} \qquad (1)$

$\displaystyle \text{Here, strength of the beam, } S = Kbh^{2}, \text{ where } K \text{ is a constant.}$

$\displaystyle \Rightarrow S = Kb(4a^{2} - b^{2}) \qquad \text{[From eq. (1)]}$

$\displaystyle \Rightarrow S = K(4a^{2}b - b^{3})$

$\displaystyle \frac{dS}{db} = K(4a^{2} - 3b^{2})$

$\displaystyle \text{For maximum or minimum value of } S, \text{ we must have } \frac{dS}{db} = 0$

$\displaystyle \Rightarrow K(4a^{2} - 3b^{2}) = 0$

$\displaystyle \Rightarrow 4a^{2} = 3b^{2}$

$\displaystyle \Rightarrow b = \frac{2a}{\sqrt{3}}$

$\displaystyle \text{Substituting the value of } b \text{ in eq. (1),}$

$\displaystyle h^{2} = 4a^{2} - \left(\frac{2a}{\sqrt{3}}\right)^{2}$

$\displaystyle \Rightarrow h^{2} = \frac{12a^{2} - 4a^{2}}{3} = \frac{8a^{2}}{3}$

$\displaystyle \Rightarrow h = \frac{2\sqrt{2}}{\sqrt{3}}\,a$

$\displaystyle \text{Now, } \frac{d^{2}S}{db^{2}} = -6Kb$

$\displaystyle \Rightarrow \frac{d^{2}S}{db^{2}} = -6K\!\left(\frac{2a}{\sqrt{3}}\right) = -\frac{12Ka}{\sqrt{3}} < 0$

$\displaystyle \text{Hence, the strength of the beam is maximum when } b = \frac{2a}{\sqrt{3}} \text{ and } h = \frac{2\sqrt{2}}{\sqrt{3}}\,a.$

$\displaystyle \textbf{Question 43: }~\text{A straight line is drawn through a given point }P(1,4). \\ \text{ Determine the least value of the sum of the intercepts on the coordinate axes.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The equation of the line passing through } (1,4) \text{ with slope } m \text{ is}$

$\displaystyle y - 4 = m(x - 1) \qquad (1)$

$\displaystyle \text{Substituting } y = 0 \text{ in eq. (1),}$

$\displaystyle 0 - 4 = m(x - 1)$

$\displaystyle \Rightarrow -4 = m(x - 1)$

$\displaystyle \Rightarrow x - 1 = -\frac{4}{m}$

$\displaystyle \Rightarrow x = \frac{m - 4}{m}$

$\displaystyle \text{Substituting } x = 0 \text{ in eq. (1),}$

$\displaystyle y - 4 = m(0 - 1)$

$\displaystyle \Rightarrow y = -m + 4$

$\displaystyle \text{Hence, the } x\text{-intercept is } \frac{m - 4}{m} \text{ and the } y\text{-intercept is } 4 - m.$

$\displaystyle \text{Let } S \text{ be the sum of the intercepts. Then,}$

$\displaystyle S = \frac{m - 4}{m} + (4 - m)$

$\displaystyle \Rightarrow S = 5 - m - \frac{4}{m}$

$\displaystyle \frac{dS}{dm} = \frac{4}{m^{2}} - 1$

$\displaystyle \text{For maximum or minimum value of } S, \text{ we must have } \frac{dS}{dm} = 0$

$\displaystyle \Rightarrow \frac{4}{m^{2}} - 1 = 0$

$\displaystyle \Rightarrow m^{2} = 4$

$\displaystyle \Rightarrow m = \pm 2$

$\displaystyle \text{Now, } \frac{d^{2}S}{dm^{2}} = -\frac{8}{m^{3}}$

$\displaystyle \left(\frac{d^{2}S}{dm^{2}}\right)_{m=2} = -\frac{8}{2^{3}} = -1 < 0$

$\displaystyle \Rightarrow \text{The sum } S \text{ is maximum at } m = 2.$

$\displaystyle \left(\frac{d^{2}S}{dm^{2}}\right)_{m=-2} = -\frac{8}{(-2)^{3}} = 1 > 0$

$\displaystyle \Rightarrow \text{The sum } S \text{ is minimum at } m = -2.$

$\displaystyle \text{Thus, the minimum value of } S \text{ is}$

$\displaystyle S = \frac{-2 - 4}{-2} + (4 - (-2)) = 3 + 6 = 9.$

$\displaystyle \textbf{Question 44: }~\text{The total area of a page is }150\text{ cm}^2.\text{ The combined } \\ \text{width of the margins at the top and bottom is }3\text{ cm and the side margin is }2\text{ cm. } \\ \text{What must be the dimensions of the page so that the area of the printed matter } \\ \text{may be maximum?}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Let } x \text{ and } y \text{ be the length and breadth of the rectangular page respectively.}$

$\displaystyle \text{Area of the page } = 150$

$\displaystyle \Rightarrow xy = 150$

$\displaystyle \Rightarrow y = \frac{150}{x} \qquad (1)$

$\displaystyle \text{Area of the printed matter } = (x - 3)(y - 2)$

$\displaystyle \Rightarrow A = xy - 2x - 3y + 6$

$\displaystyle \Rightarrow A = 150 - 2x - \frac{450}{x} + 6$

$\displaystyle \Rightarrow A = 156 - 2x - \frac{450}{x}$

$\displaystyle \frac{dA}{dx} = -2 + \frac{450}{x^{2}}$

$\displaystyle \text{For maximum or minimum value of } A, \text{ we must have } \frac{dA}{dx} = 0$

$\displaystyle \Rightarrow -2 + \frac{450}{x^{2}} = 0$

$\displaystyle \Rightarrow 2x^{2} = 450$

$\displaystyle \Rightarrow x^{2} = 225$

$\displaystyle \Rightarrow x = 15$

$\displaystyle \text{Substituting the value of } x \text{ in eq. (1),}$

$\displaystyle y = \frac{150}{15} = 10$

$\displaystyle \text{Now, } \frac{d^{2}A}{dx^{2}} = -\frac{900}{x^{3}}$

$\displaystyle \Rightarrow \left(\frac{d^{2}A}{dx^{2}}\right)_{x=15} = -\frac{900}{15^{3}} = -\frac{900}{3375} < 0$

$\displaystyle \text{Hence, the area of the printed matter is maximum when } x = 15 \text{ and } y = 10.$

$\displaystyle \textbf{Question 45: }~\text{The space }s\text{ described in time }t\text{ by a particle moving in } \\ \text{a straight line is given by }s=t^5-40t^3+30t^2+80t-250.\text{ Find the minimum } \\ \text{value of acceleration.}$
$\displaystyle \text{Answer:}$

$\displaystyle \text{Given: } s = t^{5} - 40t^{3} + 30t^{2} + 80t - 250$

$\displaystyle \Rightarrow \frac{ds}{dt} = 5t^{4} - 120t^{2} + 60t + 80$

$\displaystyle \text{Acceleration, } a = \frac{d^{2}s}{dt^{2}} = 20t^{3} - 240t + 60$

$\displaystyle \Rightarrow \frac{da}{dt} = 60t^{2} - 240$

$\displaystyle \text{For maximum or minimum value of } a, \text{ we must have } \frac{da}{dt} = 0$

$\displaystyle \Rightarrow 60t^{2} - 240 = 0$

$\displaystyle \Rightarrow 60t^{2} = 240$

$\displaystyle \Rightarrow t^{2} = 4$

$\displaystyle \Rightarrow t = 2 \quad (\text{taking } t>0)$

$\displaystyle \text{Now, } \frac{d^{2}a}{dt^{2}} = 120t$

$\displaystyle \Rightarrow \left(\frac{d^{2}a}{dt^{2}}\right)_{t=2} = 240 > 0$

$\displaystyle \text{Hence, acceleration is minimum at } t = 2.$

$\displaystyle a_{\min} = 20(2)^{3} - 240(2) + 60$

$\displaystyle \Rightarrow a_{\min} = 160 - 480 + 60 = -260$

$\displaystyle \therefore \text{At } t = 2,\; a = -260.$

$\displaystyle \textbf{Question 46: }~\text{A particle is moving in a straight line such that its distance }s\text{ at } \\ \text{any time }t\text{ is given by }s=\dfrac{t^4}{4}-2t^3+4t^2-7.\text{ Find when its velocity } \\ \text{is maximum and acceleration minimum.}$
$\displaystyle \text{Answer:}$