#### Answer

Length of plot $=12.5yd$
Width of plot $=12.5yd$
Maximum area of plot $=156.25y{{d}^{2}}$

#### Work Step by Step

Let us assume that the length of the plot be $x$ and breadth of the plot along the river be $y$.
The plot is to be fenced from three sides and is to be left along the riverside. The total available length of fence is $50$ yards.
$2x+2y=50$.
Solve for $y$.
$\begin{align}
& 2y=50-2x \\
& y=25-x.
\end{align}$
Area of the plot, $A$ , can be written as
$A=xy$.
Put $y=25-x$.
$\begin{align}
& A=x\left( 25-x \right) \\
& =-{{x}^{2}}+25x
\end{align}$.
And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-1,\ b=25$
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:
$\begin{align}
& -\frac{b}{2a}=-\left( \frac{25}{-2} \right) \\
& =12.5.
\end{align}$
Put $x=12.5$ in equation for $y$.
$\begin{align}
& y=25-x \\
& =25-12.5 \\
& =12.5.
\end{align}$
Hence, for the area of the plot to be maximum the length of the plot would be $12.5$ yards and width would be $12.5$ yards.
The area of the plot would be
$\begin{align}
& A=xy \\
& =\left( 12.5\times 12.5 \right) \\
& =156.25y{{d}^{2}}.
\end{align}$