# Vanessa has 180 feet of fencing that she intends to use to build a rectangular play area for her dog. She wants the play area to enclose at least 1800 square feet. What are the possible widths of the play area?

Nov 12, 2017

The possible widths of the play area are : 30 ft. or 60 ft.

#### Explanation:

Let length be $l$ and width be $w$

Perimeter = $180 f t . = 2 \left(l + w\right)$---------(1)

and

Area = $1800 f t {.}^{2} = l \times w$----------(2)

From (1),

$2 l + 2 w = 180$

$\implies 2 l = 180 - 2 w$

$\implies l = \frac{180 - 2 w}{2}$

$\implies l = 90 - w$

Substitute this value of $l$ in (2),

$1800 = \left(90 - w\right) \times w$

$\implies 1800 = 90 w - {w}^{2}$

$\implies {w}^{2} - 90 w + 1800 = 0$

Solving this quadratic equation we have :

$\implies {w}^{2} - 30 w - 60 w + 1800 = 0$

$\implies w \left(w - 30\right) - 60 \left(w - 30\right) = 0$

$\implies \left(w - 30\right) \left(w - 60\right) = 0$

$\therefore w = 30 \mathmr{and} w = 60$

The possible widths of the play area are : 30 ft. or 60 ft.

Nov 12, 2017

$30 \text{ or "60" feet}$

#### Explanation:

$\text{using the following formulae related to rectangles}$

$\text{where "l" is the length and "w" the width}$

• " perimeter (P) "=2l+2w

• " area (A) "=lxxw=lw

$\text{the perimeter will be "180" feet "larrcolor(blue)"fencing}$

$\text{obtaining "l" in terms of } w$

$\Rightarrow 2 l + 2 w = 180$

$\Rightarrow 2 l = 180 - 2 w$

$\Rightarrow l = \frac{1}{2} \left(180 - 2 w\right) = 90 - w$

$A = l w = w \left(90 - w\right) = 1800$

$\Rightarrow {w}^{2} - 90 w + 1800 = 0 \leftarrow \textcolor{b l u e}{\text{quadratic equation}}$

$\text{the factors of + 1800 which sum to - 90 are - 30 and - 60}$

$\Rightarrow \left(w - 30\right) \left(w - 60\right) = 0$

$\text{equate each factor to zero and solve for } w$

$w - 30 = 0 \Rightarrow w = 30$

$w - 60 = 0 \Rightarrow w = 60$