# She has 100 feet of fencing available. what is the largest area that the yard can contain?

## Question:

She has 100 feet of fencing available. what is the largest area that the yard can contain?

## The Perimeter and Area of a Rectangle:

A rectangle is a two-dimensional figure that has its length longer than its wide. For example, a tennis court is an example of a rectangle. The perimeter of a rectangle is twice the sum of its width and length {eq}P = 2(l + w) {/eq}. The area of a rectangle is the amount of space covered by the figure, and it is calculated as {eq}A= l\times w {/eq}.

Assuming that the yard is rectangular in shape, then the perimeter of the fencing will be equal to:

• {eq}P = 2(l + w) {/eq}

If we have 100ft of fencing available, then:

• {eq}100 = 2(l + w) {/eq}
• {eq}50 = l + w {/eq}.......................................................................................................................(i)

The area of the yard will be given by:

• {eq}A = l\times w {/eq}

From equation (i):

• {eq}l = 50 - w {/eq}

Therefore:

• {eq}A = (50 - w)w {/eq}
• {eq}A = 50w - w^2 {/eq}
• {eq}A = -w^2 + 50w {/eq}

Since the area assumes a quadratic equation, the maximum area will occur where:

• {eq}w = \dfrac{-b}{2a} {/eq}

{eq}a = -1, \quad b = 50 {/eq}. Therefore:

• {eq}w = \dfrac{-50}{2(-1)} {/eq}
• {eq}w = 25\; \rm ft {/eq}

Therefore, the maximum area that can be enclosed using 100ft of the available fencing is:

• {eq}A = 50(25) - 25^2 {/eq}
• {eq}\boxed{\color{blue}{A = 625\; \rm ft^2}} {/eq}