Applied Optimization

Question

a rectangular field with a partition is to be enclosed with a fence against an existing wall, so that no fence is needed on that side. The material for the fence cost $3/ft. for the two end sides and the partition perpendicular to the wall and $5/ft. for the side parallel to the wall. Find: the dimensions of the field of largest area that can be enclosed for $360, the maximum area

Answer 1

Let x = length of the side parallel to the wall

 

     y = length of each of the other sides

 

Given:  5x + 3y + 3y + 3y = 360

 

                           5x + 9y = 360     So, y = (360 - 5x)/9

 

Maximize area of the field = A = xy = x(360 - 5x)/9

 

                                                A(x) = -(5/9)x² + 40x, 0 < x < 72

 

A'(x) = (-10/9)x + 40     A'(x) = 0 when x = 36

 

When 0<x<36, A'(x) >0.  So, A(x) is increasing

 

When 36<x<72, A'(x)<0.  So, A(x) is decreasing

 

The area is maximized when x = 36 ft and y = (360-5x)/9 = 20 ft