SOLUTION: Andy plans to build a rectangular dog pen in his backyard. A portion of the back wall of his house will serve as one side of the dog pen. If the perimeter of the dog pen must be 40

Hi there,

Your Problem:
Andy plans to build a rectangular dog pen in his backyard. A portion of the back wall of his house will 
serve as one side of the dog pen. If the perimeter of the dog pen must be 40 feet, which of the following 
widths will result in the maximum area? 

Solution:
Recall that "perimeter" means the distance around the outside of a shape. We'll start with the formula 
for the perimeter of a rectangle.

Let W be the width of the rectangle.
Let L be the length of the rectangle.

To go around the outside of a rectangle we have 2 lengths and 2 widths, so the formula for the 
perimeter is 2L + 2W.

We are told that the perimeter is 40 feet, so

2L + 2W = 40

Now rewrite this equation to give L in terms of  W. (We want L by itself on the left side.)

Subtract 2W from both sides, and simplify.
2L + 2W - 2W = 40 - 2W
2L = 40 - 2W

Divide both sides by 2, and simplify.
2W/2 = (40 - 2W)/2

L = 20 - W

The problem asks you to find the rectangle width that give the maximum area. The formula for the area 
of a rectangle is A = L * W

Substitute 20 - W for L in the formula.
A = (20 - W)*W

Now we have an equation that relates the width of the rectangle to the area.
Use the distributive property to clear the parentheses.
A = 20W - W^2

I am going to rearrange the terms on the right side.
A = -W^2 + 20W

Notice this is a quadratic equation. Its graph is a parabola. Since the W-squared term is negative, the 
parabola opens downward. We will find the maximum value for A at the vertex of the parabola.

The formula for the x-value of the vertex of a parabola is -b/2a where a and b come from the quadratic 
equation in standard form, y = ax^2 + bx + c

For our area equation we have A for y and W for x, or
A = aW^2 + bW + C

a = -1  (because -W^2)
b = 20 (because 20W)
c = 0 (because no constant term)

So, the -b/2a = -(20)/2(-1) = 10

The vertex of this parabola is at W = 10.

In the context of this problem W = 10 means that a width of 10 feet with give the maximum area for the 
fenced in area.

There is an interesting thing about this type of problem. Notice that when W equals 10 feet, L also equals 
10 feet because L = 20-W. (20-10 = 10)

This means that the rectangle that gives the maximum area is a 10 x 10 square. When you have a fixed 
perimeter, a square will always give the maximum area.

Hope this helps. Feel free to email me if you have questions about this solution.

~Mrs. Figgy
math.in.the.vortex@gmail.com