- #1

Monkey618

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## Homework Statement

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?

## Homework Equations

Volume of cylinder = πr²h

Surface area of cylinder = 2πr² + 2πrh

## 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.