# What dimensions will produce the greatest area for Sharon's puppy to play, if she purchased 40 feet of fencing to enclose three sides of a fence?

May 20, 2018

If the shape is a rectangle, the area will be $200 s q f t$

#### Explanation:

The fencing is to be used for $3$ sides, If we assume that the fourth side is a wall or an existing fence, then the shape is a rectangle.

Let the length of each of the shorter sides (the breadth) be $x$.
The length will be $40 - 2 x$

$A = x \left(40 - 2 x\right)$

$A = 40 x - 2 {x}^{2}$

For a maximum, $\frac{\mathrm{dA}}{\mathrm{dx}} = 0$

$\frac{\mathrm{dA}}{\mathrm{dx}} = 40 - 4 x = 0$

$\text{ } x = 10$

The dimensions will be $10 \times 20$ feet, giving an area of $200 s q f t .$

If the shape is to be an equilateral triangle:

A = 1/2 ab sin60° = 1/2 xx40/3 xx40/3 xxsin60

$A = 76.9 s q f t$ which is much smaller than a rectangle.

If the fencing is used to form a semi-circle against a wall, the area will be:

$r = \frac{C}{2 \pi} = 80 \left(2 \pi\right) = 12.732$ feet

$A = \pi {r}^{2} = {12.732}^{2} = 162 s q f t$

May 20, 2018

Using a quadratic to solve this question.

So the length of the side is $10 \text{ feet.}$
So the length of the front is $40 - 2 \left(10\right) = 20 \text{ feet.}$

The maximum area is $20 \times 10 = 200 {\text{ feet}}^{2}$

#### Explanation:

The wording: to enclose 3 sides of a fence implies there is at least one more side.

Assumption: The shape is that of a rectangle.

Set area as $A$
Set length of front as $F$
Set length of side as $S$

Given: $F + 2 S = 40 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

Known: $A = F \times S \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(2\right)$

From $E q n \left(1\right)$ we have $F = 40 - 2 S \text{ } \ldots . E q u a t i o n \left({1}_{a}\right)$

Using $E q n \left({1}_{a}\right)$ substitute for $F$ in $E q n \left(2\right)$

$\textcolor{g r e e n}{A = \textcolor{red}{F} \times S \textcolor{w h i t e}{\text{dddd")->color(white)("dddd}} A = \textcolor{red}{\left(- 2 S + 40\right)} \times S}$

$\textcolor{g r e e n}{\textcolor{w h i t e}{\text{ddddddddddddd")->color(white)("dddd}} A = - 2 {S}^{2} + 40 S}$

This is a quadratic of general shape $\bigcap$ as the squared term is negative. Thus there is a maximum value of $A$ and it is at the vertex.

$\textcolor{b r o w n}{\text{A very useful trick to find the vertex}}$

Using the beginnings of completing the square write as:

$A = - 2 \left({S}^{2} \textcolor{red}{- \frac{40}{2}} S\right)$

${S}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \textcolor{red}{- \frac{40}{2}} = + 10$

So the length of the side is $10 \text{ feet.}$
So the length of the front is $40 - 2 \left(10\right) = 20 \text{ feet.}$

The maximum area is $20 \times 10 = 200 {\text{ feet}}^{2}$