#### Answer

Length of plot $=50ft$
Width of plot $=100ft$
Maximum area of plot $=5,000f{{t}^{2}}$

#### 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 $200ft$.
$2x+y=200$.
Solve for $y$.
$y=200-2x$.
Area of the plot, $A$ , can be written as
$A=xy$.
Put $y=200-2x$.
$\begin{align}
& A=x\left( 200-2x \right) \\
& =-2{{x}^{2}}+200x
\end{align}$.
And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-2,b=200$.
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:
$\begin{align}
& -\frac{b}{2a}=-\left( \frac{200}{-4} \right) \\
& =50
\end{align}$.
Put $x=50$ in equation for $y$.
$\begin{align}
& y=200-2x \\
& =200-2\left( 50 \right) \\
& =100.
\end{align}$
Hence, for the area of the plot to be maximum the length of the plot would be $100ft$ and width would be $50ft$.
The area of the plot would be
$\begin{align}
& A=xy \\
& =\left( 50\times 100 \right)f{{t}^{2}} \\
& =5,000f{{t}^{2}}.
\end{align}$