General Contents

More Max-Min Problems

Index

Maximize area of 3 pens with a fixed length of fencing:

If you find this page helpful and would recommend that I create more pages like this one, please let me know: Email to John Taylor


1200 ft of fencing is available to create the three pens shown. What dimensions, x and y should be used so as to enclose the maximum area? What is the maximum area?

To solve this problem, we need an expression for the area in terms of one variable. Then we can use the first derivative to find the value of the variable which yields the maximum area.

First we must express the area in terms of the 2 unknowns:

area in terms of the two dimensions

Now we must relate the dimensions to the fixed amount of fencing:

length of fencing related to the symbolic dimensions

We can combine these equations to express the area in terms of a single variable, x. Rewriting the fencing equation, we get

.

Now substitute in the equation for area and get

area in terms of one variable

Now we find the derivative:

derivative of the area

If we set the derivative to 0, the condition at a maximum, we can solve for the x dimension, xMax:

, or the value of a dimension which maximizes the area of the pens

If we substitute this value in the above equation for y, we can obtain the corresponding yMax:

the value of the other dimension which maximizes the area of the pens

We can find the maximum area by using these two dimensions in the original area equation:

the maximum area of the three pens

We can check our result by trying values of x and y which depart from these slightly:

x

90

110

y

165

135

A

44550

44550

 

The table reveals that xMax and yMax indeed result in a maximum area.

General Contents

More Max-Min Problems

Index