I am not sure how to go about solving this problem. Any help would be much appreciated.

Question

16-oz cans of soup require a volume of 36 cubic inches. Materials for the can cost t $0.005 per square inch of surface area, determine minimum cost to construct a can that holds 36 cubic inches.

Answer 1

This is an optimization problem in differential calculus.   For a can of height, h, and radius, r, the formulas for volume and surface area are:

 

  V=   pi r² h

 

  A =  2 pi r h +  pi r² + pi r²      (side plus top plus bottom )

 

The minimum cost will be obtained by minimizing A.   The formula above for A depends on two variables: r and h

For this reason, A can not be directly minimized.     The way forward is to use the volume formula to develop a constraint condition.    The volume must be 36   so

 

   36 = pi r² h.       This can be rearranged to get   h =   36 /(pi r²)  

   This is the needed constraint equation.   Substituting for h in the Area equation gives

 

   A =  2pi r 36 /(pi r²) + 2 pi r²

 

 This expression now depends on just one variable:  r.    To find the optimum value of r, one differentiates this expression with respect to r and sets the result to zero"

 

  -72 / r² + 4 pi r =0       So

 

   r = (72/ 4 pi)^1/3

 

   This works out to be   1.7989     (unit is inches)

 

  Substituting this value into the second equation for A gives;

 

    A =60.355       (unit is square inches)

 Since each square inch costs  $0.005,   the (minimum) cost will be $0.302   .  Or a little more than 30 cents.