Expert: Ahmed Salami - 12/30/2009

Question
Mary is making a cylindrical drum. The drum must hold 1.8m^3 of liquid. The drum must be no more than 1.8m tall. What are the dimensions of the drum that uses the least amount of material?

Answer
Hi Ivan,
A cylinder with radius r and height h has a volume
V = πr²h
and a total surface area
A = 2πrh + πr²  (assuming the cylinder is open)
Now, V = 1.8m³
1.8 = πr²h
h = 1.8/πr²
substituting in A,
A = 2πr(1.8/πr²) + πr²
 = 3.6/r + πr²
To find the value of r that gives the minimum value of A, we find dA/dr and equate to zero.
dA/dr = -3.6/r² + 2πr
equating to zero,
-3.6/r² + 2πr = 0
2πr = 3.6/r²
r³ = 1.8/π
  = 0.573
r = 0.83
h = 1.8/πr²
 = 1.8/π(0.83)²
 = 0.83
The cylinder should have an equal radius and height of 0.83m

If it were a closed cylinder i.e with a top
A = 2πrh + 2πr²  
As before,
h = 1.8/πr²
substituting in A,
A = 2πr(1.8/πr²) + 2πr²
 = 3.6/r + 2πr²
Now,
dA/dr = -3.6/r² + 4πr
equating to zero,
-3.6/r² + 4πr = 0
4πr = 3.6/r²
r³ = 0.9/π
  = 0.287
r = 0.66
h = 1.8/πr²
 = 1.8/π(0.66)²
 = 1.32
The cylinder should have a radius of 0.66m and height (which is twice the radius) of 1.32m

Hope its clear.

Regards