# A rectangular area is to be enclosed by a wall on one side and fencing on the other three sides....

## Question:

A rectangular area is to be enclosed by a wall on one side and fencing on the other three sides. If 18 meters of fencing are used, what is the maximum area that can be enclosed?

## The Perimeter and Area of a Rectangle:

A rectangle is a two-dimensional figure that has two pairs of parallel sides. The parallel sides of a rectangle are congruent, and each of the interior angles is a right angle. The area of a rectangle is the amount of space that is covered by a rectangular figure. We determine this area by multiplying the length of the rectangular figure by its width.

## Answer and Explanation:

To determine the maximum area that can be enclosed, we will make use of the perimeter formula and the area formula.

The perimeter of a rectangle is computed as:

- {eq}P = 2l + 2w {/eq}

And the area is calculated as:

- {eq}A = lw {/eq}

One of the sides is enclosed by a wall. Therefore, the length of the three sides enclosed by a fence will be given by:

- {eq}P = l + 2w {/eq}

If the fence is 18m, then:

- {eq}18 = l + 2w {/eq}

- {eq}l = 18 - 2w {/eq}............................................................................................(i)

Substituting equation (i) into the area formula:

- {eq}A = (18 - 2w)w = 18w - 2w^2 {/eq}

Maximizing the area:

- {eq}\dfrac{d A}{d w} = 18 - 4w = 0 {/eq}

- {eq}18 - 4w = 0 {/eq}

Solving for *w*:

- {eq}w = \dfrac{18}{4} = 4.5\, m {/eq}

Therefore, the maximum area is equal to:

- {eq}A = 18(4.5) - 2(4.5)^2 = \boxed{40.5\, m^2} {/eq}

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from Geometry: High School

Chapter 8 / Lesson 7