Q. 87 - Precalculus Enhanced with Graphing Utilities

Question

Fencing: A farmer has 300 feet of fence available to enclose a 4500-square-foot region in the shape of adjoining squares, with sides of length x and y. See the figure. Find x and y.

Step-by-Step Solution

Answer

$x=60$ and $y=30$.

Step 1. Form system of equations.

The dimensions of the fence are 

Total perimeter $=4x+2y$.

Total length of fence given is 300 feet which gives

$4x+2y=300$

The area of the adjoining squares is 4500 square foot,

$x^2+y^2=4500$

Step 2. Solve the system of equations.

From first equation,

$$\begin{gathered} 4x+2y=300 \\ 2x+y=150 \\ y=150-2x \end{gathered}$$

From second equation,

$$\begin{gathered} x^2+y^2=4500 \\ {x}^2+{(150-2x)}^2=45005x^2-600x+22500=4500 \\ 5x^2-600x+18000=0 \\ {x}^2-120x+3600=0 \end{gathered}$$

Using the quadratic equation formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$$\begin{gathered} x=\frac{120\pm \sqrt{{(-120)}^2-4\left(1\right)\left(3600\right)}}{2\left(1\right)} \\ =\frac{120\pm \sqrt{14400-14400}}{2} \\ =\frac{120\pm 0}{2} \\ =\frac{120}{2} \\ =60 \end{gathered}$$

and

$$\begin{gathered} y=150-2\left(60\right) \\ =150-120 \\ =30 \end{gathered}$$