We have a half cylinder roof of radius rr and height rr mounted on top of four rectangular walls of height hh. 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?

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1 Answer
May 31, 2016

r=20/sqrt(3)=(20sqrt(3))/3

Explanation:

Let me restate the question as I understand it.
Provided the surface area of this object is 200pi, 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 6r and height h, which have total area of 6rh.
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 = 6rh+2pi r^2
Knowing this to be equal to 200pi, we can express h in terms of r:
6rh+2pir^2=200pi
r=(100pi-pir^2)/(3r) = (100pi)/(3r) - pi/3r#

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 2r^2 and height h, that is its volume is
V_1 = 2r^2h=200/3pir - 2/3pir^3

Within the roof we have half a cylinder with radius r and height r, its volume is
V_2 = 1/2pir^3

We have to maximize the function
V(r) = V_1+V_2 = 200/3pir - 2/3pir^3 + 1/2pir^3 = 200/3pir - 1/6pir^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'(r) = 200/3pi - 1/2pi r^2

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