# A wire of length 18 m is used to form a circle and a square. Find the radius of the circle and the side length of the square that produce a minimum total area?

Jul 10, 2018

• $\left\{\begin{matrix}r \approx 1.26 m \\ s \approx 2.52 m\end{matrix}\right.$

#### Explanation:

If the wire of length is split into 2 lengths:

• ${l}_{1} , {l}_{2} q \quad \text{ with } {l}_{1} + {l}_{2} = 18$

... to make circle and square respectively.

Circle

Circle will have circumference $C$:

• $C = 2 \pi r = {l}_{1} \implies r = {l}_{1} / \left(2 \pi\right)$

• $\implies {A}_{1} = \pi {r}^{2} = {l}_{1}^{2} / \left(4 \pi\right)$

Square

Square will have perimeter $p$ with side $s$:

• $p = 4 s = {l}_{2} \implies s = {l}_{2} / 4$

• $\implies {A}_{2} = {s}^{2} = {l}_{2}^{2} / 16$

Total Area

$A \left(\boldsymbol{l}\right) = {A}_{1} \left({l}_{1}\right) + {A}_{2} \left({l}_{2}\right) = {l}_{1}^{2} / \left(4 \pi\right) + {l}_{2}^{2} / 16$

$= {l}_{1}^{2} / \left(4 \pi\right) + {\left(18 - {l}_{1}\right)}^{2} / 16$

This is now in single variable. Differentiate wrt ${l}_{1}$ to optimise:

$A ' = {l}_{1} / \left(2 \pi\right) - \frac{18 - {l}_{1}}{8} = 0$

$\implies \left\{\begin{matrix}{l}_{1} = \frac{18 \pi}{4 + \pi} \\ {l}_{2} = \frac{72}{4 + \pi}\end{matrix}\right. q \quad \left\{\begin{matrix}r \approx 1.26 m \\ s \approx 2.52 m\end{matrix}\right.$

Using approximate values for ${l}_{1} , {l}_{2}$:

$A \approx 11.3 \setminus {m}^{2}$

The 2nd derivative is:

$A ' ' = \frac{1}{2 \pi} + \frac{1}{8} > 0$

So this is a minimum .