Question

A window is in the form of rectangle surmounted by a semi-circular opening. The perimeter of window is 30 M . Find the dimensions of window so that it can admit maximum light through the whole opening
or
Prove that volume of largest cone, which can be inscribed in a sphere, is 8/27 part of sphere

Answer 1

Let $x$ and $y$ be the length and breadth of the rectangle.

Radius of circle $=\frac{x}{2}$

Perimeter of window $=30m\left(\text{Given}\right)$

Therefore,

$x+2y+\pi \frac{x}{2}=30$

$\Rightarrow y=15-(\frac{1}{2}+\frac{\pi }{4})x$

Now,

Area of window $\left(A\right)=$ Area of square $+$ Area of semicircle

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

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

$\Rightarrow A=x(15-(\frac{1}{2}+\frac{\pi }{4})x)+\frac{\pi x^2}{8}$

$\Rightarrow A=15x-(\frac{1}{2}+\frac{\pi }{8})x^2$

Differentiating above equation w.r.t. $x$, we get

$\frac{dA}{dx}=15-(1+\frac{\pi }{4})x.....\left(1\right)$

Putting $\frac{dA}{dx}=0$, we have

$15-(1+\frac{\pi }{4})x=0$

$\Rightarrow x=\frac{60}{4+\pi }$

Differentiating equation $\left(2\right)$, we get

$\frac{d^{2}A}{d{x}^2}=-(1+\frac{\pi }{4})$

At $x=\frac{60}{\pi +4}$

$\frac{d^{2}A}{d{x}^2}<0$

Thus area is maximum for $x=\frac{60}{\pi +4}$.

Therefore,

$y=15-(\frac{1}{2}+\frac{\pi }{4})\left(\frac{60}{\pi +4}\right)$

$\Rightarrow y=15-\frac{2+\pi }{4}\times \frac{60}{\pi +4}$

$\Rightarrow y=\frac{15\pi +60-30-15\pi }{\pi +4}=\frac{30}{\pi +4}$

Hence for maximum area, the dimensions of window are-

Length $=\frac{60}{\pi +4}$

Breadth $=\frac{30}{\pi +4}$

Radius of semicircle part $=\frac{30}{\pi +4}$

OR

Cone of largest volume inscribed in a sphere of radius $R$.

Volume of sphere $=\frac{4}{3}\pi R^3$

Let $OC=x$

Radius of cone $=BC$

Height of cone $=h=AC$

In right-angled triangle $BOC$,

${OB}^2={BC}^2+{OC}^2$

$\Rightarrow BC=\sqrt{R^2-x^2}$

Therefore,

Radius of cone $=\sqrt{R^2-x^2}$

Height of cone $=h=R+x$

Let $V$ be the volume of cone.

$V=\frac{1}{3}\pi r^{2}h$

$\Rightarrow V=\frac{1}{3}\pi (R^2-x^2)(R+x).....\left(1\right)$

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

Differentiating above equation w.r.t. $x$, we have

$\frac{dV}{dx}=\frac{\pi }{3}(R^2-2Rx-3x^2).....\left(2\right)$

Putting $\frac{dV}{dx}=0$

$R^2-2Rx-3x^2=0$

$(R+x)(R-3x)=0$

$x=-R$ or $x=\frac{R}{3}$

$\because x\ne -ve$

$\therefore x=\frac{R}{3}$

Now differentiating equation $\left(2\right)$ w.r.t. $x$, we have

$\frac{d^{2}V}{d{x}^2}=\frac{\pi }{3}(-2R-6x)$

$\Rightarrow \frac{d^{2}V}{d{x}^2}=-\frac{\pi }{3}(2R+6\frac{R}{3})=-\frac{4\pi R}{3}<0$

Thus Volume will be maximum for $x=\frac{R}{3}$.

Therefore,

Substituting the value of $x$ in equation $\left(1\right)$, we have

$V=\frac{\pi }{3}(R^2-\frac{R^2}{9})(R+\frac{R}{3})=\frac{\pi }{3}\times \frac{32R^3}{27}=\frac{8}{27}\left(\frac{4}{3}\pi R^3\right)$

Hence proved that the volume of largest cone, which can be inscribed in a sphere, is $\frac{8}{27}$ part of sphere.