Chapter 9.5, Problem 71E

a.

To determine

To prove: That $y=350-\frac{4}{3}x$ .

Explanation of Solution

Given information:

Measure of the pastures enclosed by fencing = 1050 feet

Area of each pasture = 15,000 feet

Consider the perimeter of the pastures be P .

So, write the total perimeter of the pastures as:

  $P=4x+3y$

Substitute 1050 in the above equation.

  $\begin{array}{c}1050=4x+3y\\ \frac{\cancel{3}y}{\cancel{3}}=-\frac{4x}{3}+\frac{1050}{3}\\ y=-\frac{4}{3}x+350\\ y=350-\frac{4}{3}x\end{array}$

Hence, the give problem is proved.

b.

To determine

To calculate: The possible lengths and widths of each pasture.

Answer to Problem 71E

The required dimensions of each pasture are $x=53.94\text{feet}$ , $y=278.1\text{feet}$ or $x=208.56\text{feet}$ , $y=71.92\text{feet}$ .

Explanation of Solution

Given information:

Measure of the pastures enclosed by fencing = 1050 feet

Area of each pasture = 15,000 square feet

Consider the perimeter of the pastures be P .

So, write the total perimeter of the pastures as:

  $P=4x+3y$

Substitute 1050 in the above equation.

  $\begin{array}{c}1050=4x+3y\\ \frac{\cancel{3}y}{\cancel{3}}=-\frac{4x}{3}+\frac{1050}{3}\\ y=-\frac{4}{3}x+350\\ y=350-\frac{4}{3}x\end{array}$

Now, the area each pasture is given by 15,000 square feet.

Write the area of each pasture can be written as:

  $xy=15000$

Substitute $350-\frac{4}{3}x$ for y in the above equation.

  $\begin{array}{c}x\left(350-\frac{4}{3}x\right)=15000\\ 350x-\frac{4}{3}{x}^2=15000\\ \cancel{3}\times \frac{4}{\cancel{3}}{x}^2-3\times 350x+3\times 15000=3\times 0\\ 4x^2-1050x+45000=0\end{array}$

Solve the above equation using Quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

  $\begin{array}{c}x=\frac{-\left(-1050\right)\pm \sqrt{{\left(-1050\right)}^2-4\left(4\right)\left(45000\right)}}{2\left(4\right)}\\ x=\frac{1050\pm \sqrt{382500}}{8}\end{array}$

Solve the above expression for x .

  $\begin{array}{c}x=\frac{1050+\sqrt{382500}}{8}\\ x=\frac{1050+618.46}{8}\\ x=\frac{1668.46}{8}\\ x\approx 208.56\end{array}$

And, $\begin{array}{c}x=\frac{1050-\sqrt{382500}}{8}\\ x=\frac{1050-618.46}{8}\\ x=\frac{431.54}{8}\\ x\approx 53.94\end{array}$

So, $x=53.94,208.56$

Now, substitute 53.94 for x in the expression $y=350-\frac{4}{3}x$ .

  $\begin{array}{c}y=350-\frac{4}{3}\left(53.94\right)\\ =350-\frac{215.76}{3}\\ =350-71.92\\ \approx 278.1\end{array}$

Also, substitute 208.56 for x in the expression $y=350-\frac{4}{3}x$ .

  $\begin{array}{c}y=350-\frac{4}{3}\left(208.56\right)\\ =350-\frac{834.24}{3}\\ =350-278.08\\ =71.92\end{array}$

Hence, the required dimensions of each pasture are $x=53.94\text{feet}$ , $y=278.1\text{feet}$ or $x=208.56\text{feet}$ , $y=71.92\text{feet}$ .