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Question 7

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

Solution

Let

• $r=$ radius of the circle

• $a=$ side of the square

Given:

Sum of the perimeters is constant:

$$2\pi r+4a=k$$

From this,

$$4a=k-2\pi r$$

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

Total Area

$$A=A_{\text{circle}}+A_{\text{square}}=\pi r^2+a^2$$

Substitute value of $a$:

$$A=\pi r^2+{\left(\frac{k-2\pi r}{4}\right)}^2$$
$$A=\pi r^2+\frac{(k-2\pi r{)}^2}{16}$$

Expand the square:

$$A=\pi r^2+\frac{k^2-4\pi kr+4{\pi }^2{r}^2}{16}$$

$$A=\pi r^2+\frac{k^2}{16}-\frac{\pi kr}{4}+\frac{{\pi }^2{r}^2}{4}$$

Combine like terms in $r^2$:

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

Differentiate to Find Minimum

$$A^{\mathrm{\prime }}(r)=-\frac{\pi k}{4}+2(\pi +\frac{{\pi }^2}{4})r$$

Set $A^{\mathrm{\prime }}(r)=0$:

$$-\frac{\pi k}{4}+2(\pi +\frac{{\pi }^2}{4})r=0$$

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

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

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

Simplify:

$$r=\frac{\pi k}{8\pi (1+\frac{\pi }{4})}=\frac{k}{8(1+\frac{\pi }{4})}=\frac{k}{2(4+\pi )}$$

Now substitute into equation for $a$:

$$a=\frac{k-2\pi r}{4}=\frac{k-2\pi \cdot \frac{k}{2(4+\pi )}}{4}$$

$$a=\frac{k-\frac{\pi k}{(4+\pi )}}{4}=\frac{k(1-\frac{\pi }{4+\pi })}{4}=\frac{k\left(\frac{4}{4+\pi }\right)}{4}$$

$$a=\frac{k}{4+\pi }$$

Compare $a$ and $r$

We have:

$$r=\frac{k}{2(4+\pi )},a=\frac{k}{4+\pi }$$

So:

$$a=2r$$

Conclusion

$$\boxed{{\displaystyle \text{The total area is least when the side of the square is }a=2r\mathrm{.}}}$$

Final Statement

$$\boxed{{\displaystyle \text{Sum of areas is minimum when the side of the square is double the radius of the circle.}}}$$

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