# Question #ef706

Apr 8, 2015

The optimal Area to Perimeter ratio is achieved by a circle.

Removal of any part of the circle's perimeter (circumference) to construct any other shape is counter productive.

Use the entire 36 meters for the circle (make your square $0 \times 0$ meters).

Apr 8, 2015

This minimization problem turns into a minimization on a closed interval. I haven't found a general post on the closed interval method. (Maybe the link owl will spot one.)

Let $x$ = the piece formed into a square.
Then the sides of the square are each of length $\frac{x}{3}$ and the area of the square is ${x}^{2} / 16$

The piece formed into a circle measures $36 - x$. That will be the circumference of the circle. We need the area as a function of this circumference.

For a circle, $A = \pi {r}^{2}$ and $C = 2 \pi r$ We have $C = 36 - x$, so that entails that $r = \frac{36 - x}{2 \pi}$ and $A = \pi {\left(\frac{36 - x}{2 \pi}\right)}^{2}$

The total area will be

$T = {x}^{2} / 16 + {\left(36 - x\right)}^{2} / \left(4 \pi\right)$ for $0 \le x \le 36$

Minimize the function of the interval:

$T ' = \frac{x}{8} - \frac{36 - x}{2 \pi} = \frac{\pi x - 144 + 4 x}{8 x}$

The zero of $T '$ occurs where $\pi x - 144 + 4 x = 0$

The critical numbers are $x = \frac{144}{4 + \pi}$ and $x = 0$.

Calculate the Total area for the 3 values of $x$,

$x = 0$, $x = \frac{144}{4 + \pi}$ and $x = 36$ to see where the minimum occurs.