Answer
The maximum area that Eastgate Consignments can enclose is $225\text{ f}{{\text{t}}^{2}}$; the dimensions are length $l=15\text{ ft}$ and width $w=15\text{ ft}$.
Work Step by Step
Let the length and width of the corner be l and w.
They have 30 ft of fencing.
So,
$l+w=30$
Substitute the value of w from the equation $l+w=30$,
$\begin{align}
& l+w=30 \\
& w=30-l
\end{align}$
Now, the area A of the corner is given by:
$A=lw$
Substitute the value $w=30-l$ in the formula of area,
That is,
$\begin{align}
& A=lw \\
& =l\left( 30-l \right) \\
& =30l-{{l}^{2}}
\end{align}$
Consider the function $y=30x-{{x}^{2}}$ and find the maximum value of y that will be the same result for A.
Use a TI-83 to draw the graph of the equation $y=30x-{{x}^{2}}$.
Step 1: Open the TI-83 graphing calculator.
Step 2: Press [Y=] key and enter the equations. Enter ${{\text{Y}}_{1}}=30X-{{X}^{2}}$.
Step 3: Press the [WINDOW] key and adjust the scale.
$\begin{align}
& X\min =0 \\
& X\max =30 \\
& Xscl=5
\end{align}$
And
$\begin{align}
& Y\min =0 \\
& Y\max =300 \\
& Yscl=100
\end{align}$
Step 4: Press the [TRACE] key.
The graph is shown below:
From the graph it can be observed that the maximum value of the function $y=30x-{{x}^{2}}$ is 225 when $x=15$.
