Chapter 7.7, Problem 57E

Dimensions of a Garden Arif’s backyard is 20 meters by 30 meters. He wants to put a flower garden in the middle of the backyard, leaving a strip of grass of uniform width around the flower garden. To be happy, he must have 184 square meters of grass. Under these conditions, what will the length and width of the garden be?

To determine

To calculate: The length and width of the garden if the length and width of the backyard are $30\text{ m}$ and $20\text{ m}$ respectively.

Answer to Problem 57E

Solution:

The length and width of the garden are $26\text{ m}$ and $16\text{ m}$ respectively.

Explanation of Solution

Given:

The length and width of the backyard are $30\text{ m}$ and $20\text{ m}$ respectively. Also, the area of the grass is $184{\text{ ft}}^2$.

Formula used:

the length and width of the backyard as $l$ and $b$ respectively. The backyard is rectangular. Then, the area of the backyard will be,

$A_{\text{backyard}}=lb$

And,

The quadratic formula to solve the quadratic equation $a{x}^2+bx+c$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a$, and $b$ are the coefficient of $x^2$ and $x$ respectively. Also, $c$ represents the constant term.

Calculation:

Consider the length and width of the backyard as $l$ and $b$ respectively. The backyard is rectangular. Thus. calculate the area of the backyard.

$A_{\text{backyard}}=lb$

Substitute $30\text{ m}$ for $l$, and $20\text{ m}$ for $b$.

$\begin{array}{c}{A}_{\text{backyard}}=\left(30\text{ m}\right)\left(20\text{ m}\right)\\ =600{\text{ m}}^2\end{array}$

Consider the width of the grass as $x$. Then, the length of the flower garden is taken as $30-2x$. Also, the width of the flower garden is $20-2x$. Draw the sketch of the garden and the grass.

Similarly, from Figure 1, calculate the area of the garden.

$\begin{array}{c}{A}_{\text{garden}}=\left(30-2x\right)\left(20-2x\right)\\ =30\left(20-2x\right)-2x\left(20-2x\right)\\ =600-60x-40x+4x^2\\ =600-100x+4x^2\end{array}$

Now, from Figure 1, it is clear that when the area of the garden is subtracted from the area of the backyard, then it will give the area of the grass. Calculate the area of the grass.

$A_{\text{grass}}=A_{\text{backyard}}-A_{\text{garden}}$

Substitute $600{\text{ m}}^2$ for $A_{\text{backyard}}$, $600-100x+4x^2$ for $A_{\text{garden}}$.

$\begin{array}{c}{A}_{\text{grass}}=600-\left(600-100x+4x^2\right)\\ =600-600+100x-4x^2\\ =100x-4x^2\end{array}$

Further, substitute $184{\text{ m}}^2$ for $A_{\text{grass}}$.

$184=100x-4x^2$

Write the equation in standard form and solve by factoring.

$\begin{array}{c}4x^2-100x+184=0\\ 4\left(x^2-25x+46\right)=0\\ x^2-25x+46=0\end{array}$

Use the quadratic formula to solve the quadratic equation.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here, $a$, and $b$ are the coefficient of $x^2$ and $x$ respectively. Also, $c$ represents the constant term.

Substitute 1 for $a$, $-25$ for $b$ and $46$ for $c$.

$\begin{array}{c}x=\frac{-\left(-25\right)\pm \sqrt{{\left(-25\right)}^2-4\left(1\right)\left(46\right)}}{2\times 1}\\ =\frac{25\pm \sqrt{625-184}}{2}\\ =\frac{25\pm \sqrt{441}}{2}\\ =\frac{25\pm \sqrt{21\cdot 21}}{2}\end{array}$

Further, solve.

$x=\frac{25\pm 21}{2}$

Thus, solutions of the above equation are,

$\begin{array}{c}x=\frac{25+21}{2}\\ =\frac{46}{2}\\ =23\end{array}$

Or,

$\begin{array}{c}x=\frac{25-21}{2}\\ =\frac{4}{2}\\ =2\end{array}$

Since the width of the backyard is $20\text{ m}$, therefore, the value of $x$ cannot exceed $20\text{ m}$. So, neglect $23\text{ m}$ and take the value as $2\text{ m}$. Thus, calculate the length of the garden.

$l_{\text{garden}}=30-2x$

Substitute $2\text{ m}$ for $x$.

$\begin{array}{c}{l}_{\text{garden}}=30\text{ m}-2\left(2\text{ m}\right)\\ =30\text{ m}-4\text{ m}\\ =26\text{ m}\end{array}$

Similarly, calculate the width of the garden.

$b_{\text{garden}}=20-2x$

Substitute $2\text{ m}$ for $x$.

$\begin{array}{c}{b}_{\text{garden}}=20\text{ m}-2\left(2\text{ m}\right)\\ =20\text{ m}-4\text{ m}\\ =16\text{ m}\end{array}$.