Question

Joe will build a rectangular pen for his dog. A wall will form one side of the pen. Joe has 40 ft of fencing to form the other three sides.

Joe plans to build the pen so that it has its maximum possible area.


What will be the dimensions of Joe’s dog pen?

Enter your answers in the boxes.

_____ ft by _______ft

Answer 1

material=wall+2*side and material is 40 ft

so: 40 = w + 2sw = 40 - 2s

Area = ws, using w from above we get:

A=(40-2s)s

A=40s-2s^2d

A/ds=40-4s and d2A/ds2=-4

Since d2A/ds2 is a constant negative acceleration, when dA/ds=0, A(s) is at an absolute maximum.

dA/ds=0

when

4s=40,

s=10 ft

And since w=40-2s,

w=20 ft

So the dimensions of the pen are 20 ft by 10 ft, with the 20 ft side being opposite the wall. And the maximum possible area is thus 200 ft^2

Answer 2

Area = x (40-2x)
A' = 40 - 4x = 0

x = 10

Dimensions of the pen would be 10 feet by 20 feet