# Relationships between variables, for use of optimization

1. Mar 22, 2012

### Monkey618

1. The problem statement, all variables and given/known data

Build a steel drum (right circular cylinder) of fixed volume V. The cost of the material is the same for the top and sides (same gage and same cost) and disregard waste. What is the relationship between the height of the radius that will minimize the surface area of the cylinder?

2. Relevant equations
Volume of cylinder = πr²h
Surface area of cylinder = 2πr² + 2πrh

3. The attempt at a solution

Find the volume, because you have no actual value it is simply V:
V = πr²h

Set in terms of one variable:
h = V / (πr²)

Find surface are with respect to the volume of the cylinder:
S = 2πr² + 2πrh

therefore, in terms of a single variable r:
S = 2πr² + 2πr (V / πr²)

Take the derivative with respect to r so that you can find where surface area has a minimum.
S = 2πr² + ((2πrV) / (πr²))
dS = 4πr + ( πr²(2Vπ) - 2Vπr(2πr) ) / (πr²)² * dh/dr

Set the derivative to 0 and solve. But because I have no actual values, I solved for dh/dr. Was this correct?

0 = 4πr + ( πr²(2Vπ) - 2Vπr(2πr) ) / (πr²)² * dh/dr

-4πr = + ( πr²(2Vπ) - 2Vπr(2πr) ) / (πr²)² * dh/dr

-4πr / [( πr²(2Vπ) - 2Vπr(2πr) ) / (πr²)²] = dh/dr

Simplify

dh/dr = -4πr / [( πr²(2Vπ) - 2Vπr(2πr) ) / (πr²)²]

dh/dr = -4πr / [(-2V)/(r²)]

dh/dr = (2πr³) / V

Is this a proper relation between height and radius for the minimal surface area of the afore mentioned steel drum? I tend to mess up my derivatives and such in class.

2. Mar 22, 2012

### HallsofIvy

You are aware that
$$\frac{2\pi rV}{\pi r^2}= \frac{2V}{r}$$
aren't you? That's much easier to work with.

3. Mar 23, 2012

### Monkey618

Ah. I am aware of it, yes. I did not notice notice it before taking the derivative. It would have made my life easier.

With the way that I've already done it, am I correct?

EDIT:

Okay, I reworked the problem and got this. I’m pretty sure that this is correct.

V = πr²h
S = 2 πr² + 2 πrh

h = V/(πr²)

so ::
S = 2 πr² + 2 πr(V/πr²)

S = 2 πr² + (2Vπr)/(πr²)
S = 2 πr² + (2V)/r

taking the derivative, I get:

ds/dr = 4 πr - 2V/r²

set to 0 and solve

0 = 4 πr - 2V/r²

2V/r² = 4 πr

r³ = (V) / (πr)

substitute in what V actually is:

r³ = (πr²h) / 2π

r³ = (r²h) / 2

r³/r² = (h) / 2

r = h/2

Last edited: Mar 23, 2012