**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.

**Solution:**

Let *r* be the radius of the circle and *a* be the side of the square.

Then, we have:

$2 \pi r+4 a=k$ (where $k$ is constant)

$\Rightarrow a=\frac{k-2 \pi r}{4}$

The sum of the areas of the circle and the square (*A)* is given by,

$A=\pi r^{2}+a^{2}=\pi r^{2}+\frac{(k-2 \pi r)^{2}}{16}$

$\therefore \frac{d A}{d r}=2 \pi r+\frac{2(k-2 \pi r)(-2 \pi)}{16}=2 \pi r-\frac{\pi(k-2 \pi r)}{4}$

Now,$\frac{d A}{d r}=0$

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

$8 r=k-2 \pi r$

$\Rightarrow(8+2 \pi) r=k$

$\Rightarrow r=\frac{k}{8+2 \pi}=\frac{k}{2(4+\pi)}$

Now, $\frac{d^{2} A}{d r^{2}}=2 \pi+\frac{\pi^{2}}{2}>0$

$\therefore$ When $r=\frac{k}{2(4+\pi)}, \frac{d^{2} A}{d r^{2}}>0$

$\therefore$ The sum of the areas is least when $r=\frac{k}{2(4+\pi)}$.

When $r=\frac{k}{2(4+\pi)}, a=\frac{k-2 \pi\left[\frac{k}{2(4+\pi)}\right]}{4}=\frac{k(4+\pi)-\pi k}{4(4+\pi)}=\frac{4 k}{4(4+\pi)}=\frac{k}{4+\pi}=2 r$.

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

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