College Algebra, Chapter 4, 4.4, Section 4.4, Problem 98

Suppose that a rectangular lot has an area of $5000 \text{ft}^2$, A diagonal between opposite curves is measured to be $10$ft longer than one side.
What are the dimensions of the land?



If the area $A$ of the rectangular lot $5000 \text{ft}^2$, then
$A = xy = 5000$ Equation 1
Then, by Pythagorean theorem,

$
\begin{equation}
\begin{aligned}
x^2 + y^2 &= (x+10)^2 \\
\\
y^2 &= (x+10)^2 - x62\\
\\
y &= \sqrt{(x+10)^2 - x^2} && \text{Equation 2}
\end{aligned}
\end{equation}
$

By substituting Equation 2 to Equation 1

$
\begin{equation}
\begin{aligned}
xy &= 5000 \\
\\
x\left( \sqrt{(x+10)^2 - x^2} \right) &= 5000 && \text{Square both sides}\\
\\
x^2\left[ (x+10)^2 - x^2 \right] &= 5000^2 && \text{Expand}\\
\\
x^2\left[ x^2 + 20x + 100 - x^2 \right] &= 5000^2 && \text{Combine like terms}\\
\\
x^2\left[ 2x + 100 \right] &= 5000^2 && \text{Distribute } x^2\\
\\
20x^3 + 1000x^2 &= 5000^2 && \text{Subtract } 5000^2\\
\\
20x^3 + 100x^2 - 5000^2 &= 0 && \text{Dive 20 on both sides}\\
\\
x^3 + 5x^2 - 1250000 &= 0
\end{aligned}
\end{equation}
$

If we graph the function,


Based from the graph, the function $f(x) = x^3+ x^2 -1250000$ is equal to 0 whe $x$ is approximately $105$ft. So
By using Equation 2

$
\begin{equation}
\begin{aligned}
y &= \sqrt{(x+10)^2 - x^2}\\
\\
&= \sqrt{(105+10)^2 - (105)^2}\\
\\
&= 10\sqrt{22}\text{ft} \quad \text{ or } \quad 47\text{ft}
\end{aligned}
\end{equation}
$

Thus, the dimension of the lot is $105$ by $47$ft