\displaystyle \textbf{Question 1. }\text{The function } f(x)=\sin x+\cos x,\ x\in \left(0,\frac{\pi}{2}\right);\ \text{Maximum value of } f(x) \text{ is } \ \
\displaystyle \text{(a) }1 \quad (b)\ 2 \quad (c)\ \frac{1}{\sqrt{2}} \quad (d)\ \sqrt{2} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, } f(x)=\sin x+\cos x
\displaystyle f'(x)=\cos x-\sin x
\displaystyle \text{For maximum value, } f'(x)=0
\displaystyle \Rightarrow \cos x-\sin x=0
\displaystyle \Rightarrow \cos x=\sin x
\displaystyle \Rightarrow \tan x=1
\displaystyle \Rightarrow x=\frac{\pi}{4}
\displaystyle f''(x)=-\sin x-\cos x
\displaystyle f''\left(\frac{\pi}{4}\right)=-(\sin \frac{\pi}{4}+\cos \frac{\pi}{4})=-(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}})=-\sqrt{2}<0
\displaystyle \Rightarrow f(x)\text{ is maximum at }x=\frac{\pi}{4}
\displaystyle \text{Maximum value }=f\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}
\displaystyle \text{Therefore Maximum value }=\sqrt{2}\text{ at }x=\frac{\pi}{4}
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\displaystyle \textbf{Question 2. }\text{A village chief wants to build a cuboidal water tank with square base} \\ \text{and open top to store }50{,}000\ \text{L water. } \text{Cost of sheet is Rs 60,000 per m}^2. \\ \text{Find dimensions for minimum cost and the minimum cost.} \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let dimensions of open tank be } x,y,x
\displaystyle V=x^{2}y=50
\displaystyle \Rightarrow y=\frac{50}{x^{2}}
\displaystyle S=x^{2}+4xy
\displaystyle =x^{2}+4x\cdot \frac{50}{x^{2}}=x^{2}+\frac{200}{x}
\displaystyle \frac{dS}{dx}=2x-\frac{200}{x^{2}}
\displaystyle \text{Now, } \frac{dS}{dx}=2x-\frac{200}{x^{2}}=0
\displaystyle \Rightarrow 2x=\frac{200}{x^{2}} \Rightarrow x^{3}=100
\displaystyle \Rightarrow x=(100)^{1/3}
\displaystyle \frac{d^{2}S}{dx^{2}}=2+\frac{400}{x^{3}}>0
\displaystyle \Rightarrow S \text{ is minimum at } x=(100)^{1/3}
\displaystyle y=\frac{50}{x^{2}}=\frac{50}{(100)^{2/3}}
\displaystyle \text{Minimum surface area } S=(100)^{2/3}+\frac{200}{(100)^{1/3}}
\displaystyle \text{Total minimum cost }=\text{Rs }\left[(100)^{2/3}+\frac{200}{(100)^{1/3}}\right]\times 60000
\displaystyle =\text{Rs }3877982.4
\displaystyle \text{Therefore Minimum cost }=\text{Rs }3877982.4
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\displaystyle \textbf{Question 3. }\text{A manufacturer designs a cylindrical paper weight inside a sphere} \\ \text{of radius }3\sqrt{3}\ \text{cm.}  \text{Find the maximum volume of the cylinder.} \ \ \displaystyle \text{Answer:}
\displaystyle \text{Let } V \text{ be volume of cylinder}
\displaystyle V=\pi r^{2}h \qquad \cdots(i)
\displaystyle \text{From figure, } \left(\frac{h}{2}\right)^{2}+r^{2}=(3\sqrt{3})^{2}
\displaystyle \Rightarrow \frac{h^{2}}{4}+r^{2}=27
\displaystyle \Rightarrow r^{2}=27-\frac{h^{2}}{4} \qquad \cdots(ii)
\displaystyle \text{From (i) and (ii), } V=\pi\left(27-\frac{h^{2}}{4}\right)h
\displaystyle =\pi\left(27h-\frac{h^{3}}{4}\right)
\displaystyle \frac{dV}{dh}=\pi\left(27-\frac{3h^{2}}{4}\right)
\displaystyle \text{For extrema, } \frac{dV}{dh}=0
\displaystyle \Rightarrow 27=\frac{3h^{2}}{4} \Rightarrow h^{2}=36 \Rightarrow h=6
\displaystyle \frac{d^{2}V}{dh^{2}}=\pi\left(-\frac{6h}{4}\right)<0
\displaystyle \Rightarrow V \text{ is maximum at } h=6
\displaystyle r^{2}=27-\frac{36}{4}=27-9=18
\displaystyle \Rightarrow r=3\sqrt{2}
\displaystyle V=\pi(3\sqrt{2})^{2}(6)=108\pi\text{ cm}^{3}
\displaystyle \text{Therefore Maximum volume }=108\pi\text{ cm}^{3}
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\displaystyle \textbf{Question 4. }\text{Find the maximum volume of the cylinder which can be inscribed} \\ \text{in a sphere of radius }3\sqrt{3}\ \text{cm (answer in terms of }\pi).   \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } h \text{ be height of cylinder and } R \text{ be radius of cylinder.}
\displaystyle \text{Given, } 3\sqrt{3} \text{ cm is radius of sphere.}
\displaystyle \text{If } V \text{ is volume of cylinder, then } V=\pi R^{2}h \qquad \cdots(i)
\displaystyle \text{From right angled triangle, } (3\sqrt{3})^{2}=\left(\frac{h}{2}\right)^{2}+R^{2}
\displaystyle \Rightarrow R^{2}=27-\frac{h^{2}}{4} \qquad \cdots(ii)
\displaystyle \text{Putting } R^{2} \text{ in Eq. (i), we get}
\displaystyle V=\pi\left(27-\frac{h^{2}}{4}\right)h
\displaystyle =27\pi h-\frac{\pi h^{3}}{4}
\displaystyle \frac{dV}{dh}=27\pi-\frac{3\pi h^{2}}{4},\quad \frac{d^{2}V}{dh^{2}}=-\frac{3\pi h}{2}
\displaystyle \text{For maximum, } \frac{dV}{dh}=0
\displaystyle \Rightarrow 27\pi-\frac{3\pi h^{2}}{4}=0
\displaystyle \Rightarrow h^{2}=36 \Rightarrow h=6
\displaystyle \left(\frac{d^{2}V}{dh^{2}}\right)_{h=6}=-9\pi<0
\displaystyle \therefore V \text{ is maximum when } h=6
\displaystyle R^{2}=27-\frac{36}{4}=18
\displaystyle V=\pi R^{2}h=\pi(18)(6)=108\pi \text{ cm}^{3}
\displaystyle \text{Hence, maximum volume }=108\pi \text{ cm}^{3}
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\displaystyle \textbf{Question 5. }\text{A Norman window is constructed by adjoining a semicircle on the} \\ \text{top of a rectangular window. } \text{Total perimeter is }10\ \text{m. }\ \
\displaystyle \text{(i) If length and breadth are } y \text{ and } x,\ \text{then relation is:} \ \
\displaystyle \text{(a) } y=\frac{20+(\pi-2)x}{4} \quad (b)\ y=\frac{20-(\pi+2)x}{2} \\ (c)\ y=\frac{20-(\pi+4)x}{4} \quad (d)\ y=\frac{20-(\pi+2)x}{4} \ \
\displaystyle \text{(ii) Area } A \text{ expressed in } x: \ \
\displaystyle \text{(a) }A=5x+\frac{\pi}{4}x^2-2x^2 \quad (b)\ A=5x+\frac{\pi}{8}x^2-\frac{1}{2}x^2 \\ (c)\ A=5x-\frac{\pi}{8}x^2-\frac{1}{2}x^2 \quad (d)\ A=5x-\frac{\pi}{2}x^2-\frac{1}{4}x^2 \ \
\displaystyle \text{(iii) Radius of semicircle for maximum }A: \ \
\displaystyle \text{(a) }\frac{10}{2+\pi} \quad (b)\ \frac{10}{\pi-2} \quad (c)\ \frac{10}{4+\pi} \quad (d)\ \frac{20}{4-\pi} \ \
\displaystyle \text{(iv) Length } y \text{ for maximum }A: \ \
\displaystyle \text{(a) }\frac{10}{4+\pi} \quad (b)\ \frac{10}{\pi-2} \quad (c)\ \frac{20}{4+\pi} \quad (d)\ \frac{20}{4-\pi} \ \
\displaystyle \text{Answer:}
\displaystyle \text{(d) Given, diameter of semi-circle }=x
\displaystyle \text{So, radius of semi-circle }=\frac{x}{2}
\displaystyle \text{Perimeter of window }=10\text{ m}
\displaystyle x+2y+\pi\left(\frac{x}{2}\right)=10
\displaystyle \Rightarrow 2x+4y+\pi x=20
\displaystyle \Rightarrow 4y=20-(\pi+2)x
\displaystyle \Rightarrow y=\frac{20-(\pi+2)x}{4} \qquad \cdots(i)
\displaystyle \text{(c) Area of window }=
\displaystyle \text{Area of rectangle }+\text{ Area of semi-circle}
\displaystyle A=xy+\frac{1}{2}\pi\left(\frac{x}{2}\right)^{2}
\displaystyle =x\left[\frac{20-(\pi+2)x}{4}\right]+\frac{1}{2}\cdot \frac{\pi x^{2}}{4} \quad \text{[from Eq. (i)]}
\displaystyle =5x-\frac{\pi x^{2}}{4}-\frac{1}{2}x^{2}+\frac{\pi x^{2}}{8}
\displaystyle A=5x-\frac{\pi x^{2}}{8}-\frac{1}{2}x^{2}
\displaystyle \frac{dA}{dx}=5-\frac{\pi x}{4}-x
\displaystyle \text{Putting } \frac{dA}{dx}=0
\displaystyle 0=5-\frac{\pi x}{4}-x \Rightarrow x+\frac{\pi x}{4}=5
\displaystyle \Rightarrow x\left(1+\frac{\pi}{4}\right)=5
\displaystyle \Rightarrow x=\frac{5}{1+\frac{\pi}{4}}=\frac{20}{\pi+4}
\displaystyle \frac{d^{2}A}{dx^{2}}=-1-\frac{\pi}{4}<0
\displaystyle \text{Hence, } A \text{ is maximum when } x=\frac{20}{\pi+4}
\displaystyle \text{Therefore, radius of semi-circle }=\frac{x}{2}=\frac{10}{\pi+4}
\displaystyle \text{(iv) Substituting value of } x \text{ in Eq. (i)}
\displaystyle y=\frac{20-(\pi+2)\left(\frac{20}{\pi+4}\right)}{4}
\displaystyle =\frac{5}{1}-\frac{20}{\pi+4}\cdot \frac{\pi+2}{4}
\displaystyle =5-\frac{5(\pi+2)}{\pi+4}
\displaystyle =5\left(1-\frac{\pi+2}{\pi+4}\right)
\displaystyle =5\left(\frac{\pi+4-(\pi+2)}{\pi+4}\right)
\displaystyle =5\left(\frac{2}{\pi+4}\right)
\displaystyle y=\frac{10}{\pi+4}
\\

\displaystyle \textbf{Question 6. }\text{A cone is inscribed in a sphere of radius }12\ \text{cm. If volume is} \\ \text{maximum, find its height. } \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let radius of cone }=x \text{ cm and height }=h \text{ cm}
\displaystyle \text{Here, } AD=h,\ OA=OC=12\text{ cm}
\displaystyle DC=x,\ OD=h-12
\displaystyle \text{In } \triangle ODC,
\displaystyle OD^{2}+DC^{2}=OC^{2}
\displaystyle \Rightarrow (h-12)^{2}+x^{2}=12^{2}
\displaystyle \Rightarrow x^{2}=12^{2}-(h-12)^{2}
\displaystyle \Rightarrow x^{2}=24h-h^{2} \qquad \cdots(i)
\displaystyle \text{Volume of cone } V=\frac{1}{3}\pi x^{2}h
\displaystyle =\frac{1}{3}\pi (24h-h^{2})h \quad \text{[from Eq. (i)]}
\displaystyle =\frac{1}{3}\pi (24h^{2}-h^{3})
\displaystyle \frac{dV}{dh}=\frac{1}{3}\pi [48h-3h^{2}]
\displaystyle =\pi (16h-h^{2})
\displaystyle \text{For maximum, } \frac{dV}{dh}=0
\displaystyle \Rightarrow \pi(16h-h^{2})=0
\displaystyle \Rightarrow h(16-h)=0
\displaystyle \Rightarrow h=0,16
\displaystyle \text{Reject } h=0 \Rightarrow h=16
\displaystyle \frac{d^{2}V}{dh^{2}}=\pi(16-2h)
\displaystyle \left(\frac{d^{2}V}{dh^{2}}\right)_{h=16}=\pi(16-32)=-16\pi<0
\displaystyle \therefore V \text{ is maximum at } h=16
\displaystyle \text{Again, on differentiating w.r.t. } h,\ \text{we obtain}
\displaystyle \frac{d^{2}V}{dh^{2}}=\pi(16-2h)
\displaystyle \left(\frac{d^{2}V}{dh^{2}}\right)_{h=16}=\pi(16-32)=-16\pi<0
\displaystyle \text{Hence, the value is maximum at height } h=16\text{ cm.}
\\

\displaystyle \textbf{Question 7. }\text{Show that altitude of a right circular cone of maximum volume} \\ \text{inscribed in sphere of radius }r\text{ is }\frac{4r}{3}.  \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let } R \text{ be radius and } h \text{ height of cone inscribed in sphere of radius } r
\displaystyle OA=h-r
\displaystyle \text{In } \triangle OAB,
\displaystyle r^{2}=R^{2}+(h-r)^{2}
\displaystyle \Rightarrow r^{2}=R^{2}+h^{2}+r^{2}-2hr
\displaystyle \Rightarrow R^{2}=2hr-h^{2} \qquad \cdots(i)
\displaystyle \text{Volume of sphere }=\frac{4}{3}\pi r^{3}
\displaystyle \text{Volume of cone }=\frac{1}{3}\pi R^{2}h
\displaystyle V=\frac{1}{3}\pi h(2rh-h^{2})
\displaystyle =\frac{1}{3}\pi(2rh^{2}-h^{3}) \qquad \cdots(ii)
\displaystyle \text{On differentiating Eq. (ii) w.r.t. } h,
\displaystyle \frac{dV}{dh}=\frac{1}{3}\pi(4rh-3h^{2}) \qquad \cdots(iii)
\displaystyle \text{For maxima/minima, } \frac{dV}{dh}=0
\displaystyle \Rightarrow 4rh-3h^{2}=0
\displaystyle \Rightarrow h(4r-3h)=0
\displaystyle \Rightarrow h=\frac{4r}{3} \quad [\because h\neq 0]
\displaystyle \text{Again, differentiating Eq. (iii),}
\displaystyle \frac{d^{2}V}{dh^{2}}=\frac{1}{3}\pi(4r-6h)
\displaystyle \text{At } h=\frac{4r}{3},
\displaystyle \frac{d^{2}V}{dh^{2}}=\frac{1}{3}\pi\left(4r-6\cdot \frac{4r}{3}\right)
\displaystyle =\frac{1}{3}\pi(4r-8r)=-\frac{4\pi r}{3}<0
\displaystyle \therefore V \text{ is maximum when } h=\frac{4r}{3}.
\\

\displaystyle \textbf{Question 8. }\text{An open top box is made by removing equal squares from each} \\ \text{corner of a }3\text{ m by }8\text{ m sheet. }  \text{Find volume of largest box.  } \ \
\displaystyle \text{Answer:}
\displaystyle \text{Let side of square piece cut be } x \text{ m}
\displaystyle \text{Length of box }=(8-2x)\text{ m}
\displaystyle \text{Breadth of box }=(3-2x)\text{ m}
\displaystyle \text{Height of box }=x
\displaystyle V=lbh=(8-2x)(3-2x)x
\displaystyle =x(24-22x+4x^{2})
\displaystyle =24x-22x^{2}+4x^{3}
\displaystyle \frac{dV}{dx}=24-44x+12x^{2}
\displaystyle \text{For maxima/minima, } \frac{dV}{dx}=0
\displaystyle 24-44x+12x^{2}=0
\displaystyle \Rightarrow 3x^{2}-11x+6=0
\displaystyle \Rightarrow (x-3)(3x-2)=0
\displaystyle \Rightarrow x=3,\ \frac{2}{3}
\displaystyle \text{But } x=3 \text{ is not feasible.}
\displaystyle \text{Again, } \frac{d^{2}V}{dx^{2}}=-44+24x
\displaystyle \text{At } x=\frac{2}{3},\ \frac{d^{2}V}{dx^{2}}=-44+16=-28<0
\displaystyle \therefore V \text{ is maximum at } x=\frac{2}{3}\text{ m}
\displaystyle \text{Maximum volume }=(8-\frac{4}{3})(3-\frac{4}{3})\cdot \frac{2}{3}
\displaystyle =\frac{20}{3}\cdot \frac{5}{3}\cdot \frac{2}{3}
\displaystyle =\frac{200}{27}\text{ m}^{3}
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