Question

The sum of the perimeter of a circle and square is k, where k is some constant, Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Answer 1

Given, $2\pi r+4x=k\Rightarrow x=\frac{k-2\pi r}{4}$ ...(i)
$\therefore A=x^2+\pi r^2=[\frac{k-2\pi r}{4}{]}^2+\pi r^2=(\frac{1}{16}\left)\right(k^2-4k\pi r+4{\pi }^2{r}^2)+\pi r^2$
On differentiating w.r.t.r, we get $\frac{dA}{dr}=\left(\frac{1}{16}\right)(-4k\pi +8{\pi }^{2}r)+2\pi r$
Again, differentiating w.r.t.r, we get
$\frac{d^{2}A}{d{r}^2}=\frac{1}{16}[0+8{\pi }^2]+2\pi =2\pi +\frac{{\pi }^2}{2}>0$
For maximum or minimum, put $\frac{dA}{dr}=0$
$\Rightarrow 2\pi r-\frac{4k\pi }{16}+\frac{8{\pi }^{2}r}{16}=0\Rightarrow r(2\pi +\frac{{\pi }^2}{2})=\frac{k\pi }{4}\Rightarrow r=\frac{\left(\frac{k\pi }{4}\right)}{2\pi +\frac{{\pi }^2}{2}}=\frac{k}{8+2\pi }$ ...(ii)
Now, $(\frac{d^{2}A}{d{r}^2}{)}_{r=\frac{k}{8+2\pi }}$ = Positve
$\therefore$ A is least when $r=\frac{k}{8+2\pi }$ and put this value in Eq. (i), we get
$x=\frac{k-2\pi r}{4}=\frac{1}{4}(k-2\pi \times \frac{k}{8+2\pi })=\frac{1}{4}\left[\frac{8k+2\pi k-2\pi k}{8+2\pi }\right]$
$=\frac{2k}{8+2\pi }=2\left(\frac{k}{8+2\pi }\right)=2r$ [Using Eq.(ii)]
Hence, S is least when side of the square is double the radius of the circle.