Chapter 1.7, Problem 67PS

Solution

a.

To write: An equation for the area of the eating section.

A equation is $x\left(120-2x\right)=1500$ .

Given:

School is adding a rectangular outdoor eating section along part of a $70$ foot side of the school. The eating section will be enclosed by a fence along its three open sides.

The school has $120$ feet of fencing and plans to use $1500$ square feet of land for the eating section.

  

Calculation:

The eating section is in the shape of a rectangle. The area of a rectangle is the product of its length and width.

From the figure, find that the length is $x$ , and the width is $120-2x$ . The area of the section is given as $1500$ square feet.

Thus, an equation for the area of the eating section is $x\left(120-2x\right)=1500$

b.

To find: Solution of an equation.

Solution of an equation is $42.25$ .

Given:

School is adding a rectangular outdoor eating section along part of a $70$ foot side of the school. The eating section will be enclosed by a fence along its three open sides.

The school has $120$ feet of fencing and plans to use $1500$ square feet of land for the eating section.

  

Calculation:

For solving the equation, first apply the distributive property.

  $\begin{array}{r}\hfill 120x-2x^2=1500\\ \hfill -2x^2+120x=1500\end{array}$

Divide each term by $-2$ .

  $\begin{array}{c}\frac{-2x^2}{-2}+\frac{120x}{-2}=\frac{1500}{-2}\\ x^2-60x=-750\end{array}$

Square half the coefficient of $x$ .

  $\begin{array}{c}{\left(-\frac{60}{2}\right)}^2={\left(-30\right)}^2\\ =900\end{array}$

Add $900$ to each side of the equation.

  $\begin{array}{c}{x}^2-60x+900=-750+900\\ x^2-60x+900=150\end{array}$

Write the left side as a binomial squared and simplify.

  ${\left(x-30\right)}^2=150$

Take the square root on each side.

  $x-30=\pm \sqrt{150}$

Add $30$ to each side to solve for $x$ .

  $\begin{array}{c}x-30+30=\pm \sqrt{150}+30\\ x=30\pm \sqrt{150}\end{array}$

Simplify the radical.

  $\begin{array}{c}x=30\pm \sqrt{25\cdot 6}\\ x=30\pm 5\sqrt{6}\end{array}$

Use a calculator to evaluate.

  $x\approx 42.25\text{ or }x\approx 17.75$

The solutions are $42.25$ and $17.75$ .

Evaluate $120-2x$ for $x=42.25$ and $x=17.75$ .

  $\begin{array}{c}120-2x=120-2\left(42.25\right)\\ =120-84.5\\ =35.5\end{array}$

Also,

  $\begin{array}{c}120-2x=120-2\left(17.75\right)\\ =120-35.5\\ =84.5\end{array}$

Since the width of the eating section cannot be greater than the length of the side of the school, $84.5\text{ft}$ is not a possible width. Thus, the solution $17.75$ must be rejected.

c.

To find: Dimensions of the eating section.

The dimensions of the eating section are $42.25\text{ft}$ by $35.5\text{ft}$ .

Given:

School is adding a rectangular outdoor eating section along part of a $70$ foot side of the school. The eating section will be enclosed by a fence along its three open sides.

The school has $120$ feet of fencing and plans to use $1500$ square feet of land for the eating section.

  

Calculation:

In this case, $x$ represents the length and $120-2x$ represents the width of the eating section. The possible value for $x$ is $42.25$ and the corresponding value for $120-2x$ is $35.5$ . Therefore, the dimensions of the eating section are $42.25\text{ft}$ by $35.5\text{ft}$ .