Optimization question

Question

A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 10 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area. (Round your answer to three decimal places.)

Answer 1

Let

r be a variable denoting the radius of the cylinder,

h be a variable denoting the height of the cylinder,

S(r, h) be the surface of the solid,

V = 10 cm³ be the volume of the solid.

The volume of the solid is the sum of the volume of the cylinder,

plus the volume of a sphere of radius r.

Therefore we have the constraint

V = πr²h + 4πr³/3

h = (V - 4πr³/3) / (πr²) (1)

The surface S(r, h) of the solid consists of the side of the cylinder,

plus the surface of a sphere

S(r, h) = 2πrh + 4πr²

Substituting h from (1)

S(r) = 2πr(V - 4πr³/3) / (πr²) + 4πr²

= 2V/r + 4πr²/3

We are to find the minimum of S(r).

That occurs either at a local minimum, or at a bound of valid interval.

From physical considerations and from equation (1) we see that as r increases, h decreases.

Therefore the two bounds of the valid interval are when r = 0 or when h = 0.

Minimum cannot occur at r = 0, because as r -> 0, S(r) goes to infinity.

We will talk about the other bound h = 0 later.

First let's compute local minima by setting the derivative S'(r) to 0.

-2V/r² + 8πr/3 = 0

r³ = 3V / 4π (2)

Notice that for the above value of r

4πr³/3 = V

In other words, the local minimum (2) has the property that

the entire volume V is contained in the sphere of radius r.

In other words, the local minimum occurs at the bound of the valid interval when h = 0.

You can confirm that by plugging (2) into (1) and getting h = 0.

We could have come to that conclusion without any calculation,

because sphere is known to be the most "efficient" solid --

capable of enclosing maximum volume with minimum surface.

To get the actual numerical value for r, we plug given values into (2)

r = (3 x 10cm³ / 4π)^1/3 = 1.336 cm