# Need an answer?

## Harold Goldstein and his wife Elaine recently installed a​ built-in rectangular swimming pool measuring 23 feet by 47 feet. They want to add a decorative tile border of uniform width around all sides of the pool. How wide can they make the tile border if they purchased enough tile to cover 296 square​ feet?

Apr 4, 2018

See below.

#### Explanation:

The pool is 23ft x 47 ft.

That makes the perimeter $2 \cdot 23 + 2 \cdot 47 = 140$ ft

Let the tile border width be $x$ ft

So you have :

Area of border = $296 = 140 \cdot x$

So $x = \frac{296}{140} = 2.1$ ft

Tiles come in standard sizes, You are unlikely to find a 2.1ft ( 25.37 inches) wide tile,

So they will have to decide tile size and how much is likleyto go waste.

Apr 4, 2018

The tile border can be 2 feet wide

#### Explanation:

I modeled this as two rectangles. The inner one is the pool, and the outer one is the area of the border. If you take the difference in the areas of the rectangle you get the area coverage of the border:

${B}_{H} \times {B}_{W} - {P}_{H} \times {P}_{W} = \text{Border Area}$

${B}_{H} \cdot {B}_{W} - 47 \cdot 23 = 296$

Where ${B}_{H}$ and ${B}_{W}$ are the outer heights and widths of the border, and ${P}_{H}$/${P}_{W}$ are the pool height and width.

The outer length and width is equal to the inner length width increased by twice the border thickness, since its the same thickness on each side.

${B}_{H} = {P}_{H} + 2 t = 47 + 2 t$
${B}_{W} = {P}_{W} + 2 t = 23 + 2 t$

Where $t$ is the border thickness

Now, we substitute our solutions for ${B}_{H}$ and ${B}_{W}$ in terms of $t$:

$\left(47 + 2 t\right) \left(23 + 2 t\right) - 47 \cdot 23 = 296$

$\left(4 {t}^{2} + 140 t + 1081\right) - 1081 = 296$

$4 {t}^{2} + 140 t = 296$

$\frac{\cancel{4} {t}^{2}}{\cancel{\textcolor{red}{4}}} + \frac{140 t}{\textcolor{red}{4}} = \frac{296}{\textcolor{red}{4}}$

${t}^{2} + 35 t \textcolor{red}{- 74} = \cancel{74 \textcolor{red}{- 74}}$

${t}^{2} + 35 t - 74 = 0$

Now, we have a quadratic and we can solve for $t$

$a = 1$
$b = 35$
$c = - 74$

$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$t = \frac{- 35 \pm \sqrt{{35}^{2} - 4 \left(1\right) \left(- 74\right)}}{2 \left(1\right)}$

$t = \frac{- 35 \pm \sqrt{1225 + 296}}{2}$

$t = \frac{- 35 \pm \sqrt{1521}}{2} = \frac{- 35 \pm 39}{2}$

$t = \frac{- 35 + 39}{2} = 2$

$t = \frac{- 35 - 39}{2} = - 37$

Now, we have two solutions for $t$ since it's a quadratic, but the negative solution is impossible, since there is no such thing as 'negative thickness'. This means we go with the positive root:

$\textcolor{g r e e n}{t = 2}$