Answer
Length of plot $=150ft$.
Width of plot $=300ft$.
Maximum area of plot $=45,000f{{t}^{2}}$
Work Step by Step
Let us assume that the length of the plot be $y$ and breadth of the plot along the river be $x$.
The plot is to be fenced from three sides and is to be left along the riverside. The total available length of fence is $600ft$.
$2x+y=600$.
Solve for $y$.
$y=600-2x$.
Area of the plot, $A$ , can be written as
$A=xy$.
Put $y=600-2x$.
$\begin{align}
& A=x\left( 600-2x \right) \\
& =-2{{x}^{2}}+600x
\end{align}$.
And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-2,\ b=600$.
Since $a<0$ , the given quadratic function has a maximum. The value at which the function gives a maximum value is calculated by evaluating $-\frac{b}{2a}$ which is shown as follows:
$\begin{align}
& -\frac{b}{2a}=-\left( \frac{600}{-4} \right) \\
& =150.
\end{align}$
Put $x=150$ in equation for $y$.
$\begin{align}
& y=600-2x \\
& =600-2\left( 150 \right) \\
& =300.
\end{align}$
Hence, for the area of the plot to be maximum the length of the plot would be $300ft$ and width would be $150ft$.
The area of the plot would be
$\begin{align}
& A=xy \\
& =\left( 150\times 300 \right)f{{t}^{2}} \\
& =45,000f{{t}^{2}}.
\end{align}$
