Answer
a) $1250$ square feet
b) Length $50$ feet, width $25$ feet
Work Step by Step
We are given 100 feet of lights that we would like to use as a rectangular fence around the building shown in the figure (see textbook). We want to maximize the area of the rectangle knowing that one side of the rectangle is the wall of the house. This means that the parameter of the rectangle will have 3 sides. In other words, the perimeter $P$ is given by.
\begin{equation}
\begin{aligned}
P &= l+2w= 100 \\
l & =100-2w.
\end{aligned}
\end{equation} The area of the rectangle can be written as a function $w$ as shown:
\begin{equation}
\begin{aligned}
A&=lw \\
& =\left( 100-2w \right)w \\
&= -2w^2+100w.
\end{aligned}
\end{equation} We see that the area of the rectangle is a parabola with $a= -2$. This means that the vertex is the maximum point of the parabola and so, the area is a maximum. The value of $w$ that gives the maximum area is given by: \begin{equation}
\begin{aligned}
w&=-\frac{b}{2a} \\
& =-\frac{100}{2\cdot (-2)} \\
&= 25.
\end{aligned}
\end{equation} A) The maximum area of the rectangle is:
\begin{equation}
\begin{aligned}
A_{max}&=-2\cdot(25)^2+100\cdot (25) \\
&= 1250.
\end{aligned}
\end{equation} The maximum area of the rectangle is $1250$ square feet.
B) Find the value of the long side of the rectangle. \begin{equation}
\begin{aligned}
l & =100-2(25) \\
&= 50.
\end{aligned}
\end{equation} The length and the width that give the largest area of the rectangle are $50$ feet by $25$ feet.
