**Question:**

Given the sum of the perimeters of a square and a circle, show that the sum of there areas is least when one side of the square is equal to diameter of the circle.

**Solution:**

Let the length of a side of the square and radius of the circle be $x$ and $r$, respectively. It is given that the sum of the perimeters of square and circle is constant.

$\Rightarrow 4 x+2 \pi r=K$ (Where $K$ is some constant)

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

Now,

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

$\Rightarrow A=\frac{(K-2 \pi r)^{2}}{16}+\pi r^{2}$ [From eq. (1) ]

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

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

$\Rightarrow K-2 \pi r=8 r$ .....(2)

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

From eqs. (1) and (2), we get

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

$\Rightarrow x=\frac{8 r}{4}$

$\Rightarrow x=2 r$

$\therefore$ Side of the square $=$ Diameter of the circle

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