# We have a half cylinder roof of radius r and height r mounted on top of four rectangular walls of height h. We have 200π m^2 of plastic sheet to be used in the construction of this structure. What is the value of r that allows maximum volume?

May 31, 2016

$r = \frac{20}{\sqrt{3}} = \frac{20 \sqrt{3}}{3}$

#### Explanation:

Let me restate the question as I understand it.
Provided the surface area of this object is $200 \pi$, maximize the volume.

Plan
Knowing the surface area, we can represent a height $h$ as a function of radius $r$, then we can represent volume as a function of only one parameter - radius $r$.
This function needs to be maximized using $r$ as a parameter. That gives the value of $r$.

Surface area contains:
4 walls that form a side surface of a parallelepiped with a perimeter of a base $6 r$ and height $h$, which have total area of $6 r h$.
1 roof, half of a side surface of a cylinder of a radius $r$ and hight $r$, that has area of $\pi {r}^{2}$
2 sides of the roof, semicircles of a radius $r$, total area of which is $\pi {r}^{2}$.

The resulting total surface area of an object is
$S = 6 r h + 2 \pi {r}^{2}$
Knowing this to be equal to $200 \pi$, we can express $h$ in terms of $r$:
$6 r h + 2 \pi {r}^{2} = 200 \pi$
$r = \frac{100 \pi - \pi {r}^{2}}{3 r} = \frac{100 \pi}{3 r} - \frac{\pi}{3} r$#

The volume of this object has two parts: Below the roof and within the roof.

Below the roof we have a parallelepiped with area of the base $2 {r}^{2}$ and height $h$, that is its volume is
${V}_{1} = 2 {r}^{2} h = \frac{200}{3} \pi r - \frac{2}{3} \pi {r}^{3}$

Within the roof we have half a cylinder with radius $r$ and height $r$, its volume is
${V}_{2} = \frac{1}{2} \pi {r}^{3}$

We have to maximize the function
$V \left(r\right) = {V}_{1} + {V}_{2} = \frac{200}{3} \pi r - \frac{2}{3} \pi {r}^{3} + \frac{1}{2} \pi {r}^{3} = \frac{200}{3} \pi r - \frac{1}{6} \pi {r}^{3}$
that looks like this (not to scale)
graph{2x-0.6x^3 [-5.12, 5.114, -2.56, 2.56]}

This function reaches its maximum when it's derivative equals to zero for a positive argument.

$V ' \left(r\right) = \frac{200}{3} \pi - \frac{1}{2} \pi {r}^{2}$

In the area of $r > 0$ it's equal to zero when $r = \frac{20}{\sqrt{3}} = 20 \frac{\sqrt{3}}{3}$.
That is the radius that gives the largest volume, given the surface area and a shape of an object.