Calculus I Optimization

Question

A rectangular box with square base and a fixed volume V  is to be made from two types of material. The material for the top and bottom costs t  cents per square inch and the material for the sides costs s  cents per square inch. Let the length of the base be x  and the height be h  . Find x  and h  (in terms of V  , t  , s  ) so that the cost is minimal.

 

 

x=_____inches.

 

h=_____inches.

Answer 1

V = x²h so solving for h gives h = V/x².

 

The cost for the base is tx² and for each side it is shx = s(V/x²)x = sV/x.

 

Therefore the cost function is

 

C(x) = 4sV/x + tx²

 

To minimize, set C'(x) = 0 and solve for x:

 

C'(x) = -4sV/x² + 2tx

 

2tx = 4sV/x²

 

x³ = 2sV/t

 

x = ³√(2sV/t)

 

h = V/x² = ³√[Vt²/(4s²)]

 

We verify that this is minimum with the second derivative test.

 

C''(x) = 8sV/x³ + 2t

 

This is positive for all x > 0, so we do have a minimum.