optimization problem

Question

A manufacturer wishes to produce rectangular containers with square bottoms and tops, where each container has a capacity of 250 cubic inches. If the material used for the top and bottom costs twice as much per square inch as the material for the sides, what dimensions will minimize the cost of the container?

Answer 1

This is a standard problem of optimization subject to a constraint.   Initially the are two free parameters:  the height, h, of the container, and the side of the square base, w.  The constraint equation:  250 =  h w² can be used to eliminate h in favor of w in the objective function (the cost of surface materials).

 

   h =  250 / w²

 

The objective function (the cost) is  C =  4s h w  +  2 (2s) w² 

    where s is the cost per square inch of side material and 2s is the cost per square inch of top/bottom material.  This formula takes into account that there are 4 sides and 2 (top/bottom).    Using the constraint equation gives

   C  =  4 s [ 250/w  + w² ].      This is now a function of just one free parameter (w).   The optimum value of w is found by setting the derivative of C with respect to w equal to zero.  This results in the equation

   

    0 =  - 250/w²  + 2 w       rearranging  gives   125 = w³   so  w  = 5.

 

  The constraint equation then gives  h = 10