Materials for cheapest container

Question

A rectangular storage container with an open top has a volume of 10m^3. The length of the base is twice the width. Material for the base costs $10 per square meter. Material for the sides cost $6 per square meter. Find the cost of the materials for the cheapest such container

Answer 1

V= lwh

l = 2w

10 = (2w)(w)(h), so h = 5/w²

 

The cost function is found by multiplying area of base by $10, area of each of the 4 sides by $6.

 

So, C= 10(l)(w) + 6(2)(w)(h) + 6(2)(l)(h)  -> there are two sides that are wh and two that are lh.

 

C= 10(2w)(w) + 12(w)(5/w²) + 12(2w)(5/w²)

C=20w² + 60w^-1 + 120w^-1 = 20w² + 180w^-1

 

Finally C' = 40w-180w^-2

 

Find critical numbers by setting C' = 0 => 180/w² = 40w

w³=18/4

 

w=(18/4)^1/3

 

Can prove that this yields a minimum for cost function (left to the reader--perhaps 2nd derivative test?)

 

C=20(18/4)^2/3 + 180/(18/4)^1/3  (dollars)