Optimization

Question

Dianna wants to enclose a rectangular flower plot that adjoins a brick retaining wall. If she has only 30 m of decorative fencing, what should be the dimensions of the flower plot be so that she a maximum area? 

Answer 1

The plot will join a brick wall. So, the plot only has three walls that vary, two will be of equal length x.

Write the perimeter excluding the brick wall side so we have a relationship to length of fence available. 

 

Perimeter = 2x + y

Area = xy

 

use the amount of fence, 30m and rewrite perimeter to equal y.

30 = 2x + y

30 - 2x = y

 

Substitute (30-2x) in for y in Area

 

Area = x(30-2x)

A(x) = 30x - 2x²

 

Take the derivative with respect to x

 

A'(x) = 30 - 4x

 

set equal to zero and solve for side length

 

0 = 30 - 4x

4x = 30

x = 30/4

x = 7.5 m

 

So the two sides of equal length measure 7.5 meters and the third side parallel to the brick wall measures;

30 - 2(7.5) = y

30 -15 = y

15m = y

 

 

You can see that it is a maximum for area by looking at the graph or table for A(x)= 30x - 2x²

 

A(x)       x

112        7

112.5    7.5

112        8

 

Answer 2

Let y = length of side parallel to the retaining wall

     x = length of each of 2 sides perpendicular to the retaining wall

 

Then, y+2x = 30.  So, y = 30 - 2x

 

Area = A = xy

 

A = x(30 - 2x)

 

A = -2x² + 30x, 0 < x < 15

 

A' = -4x + 30

 

A' = 0 when x = 7.5

 

If 0<x<7.5, A'>0.  So A is increasing

 

If 7.5<x<15, A'<0.  So, A is decreasing

 

Therefore, we have a relative and absolute maximum when x = 7.5 m and y = 15 m.