Answer
Length of plot $\ =20yd$
Width of plot $\ =20yd$
Maximum area of plot $\ =400y{{d}^{2}}$
Work Step by Step
Let us assume 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 $80$ yards.
$2x+2y=80$.
Solve for $y$.
$\begin{align}
& 2y=80-2x \\
& y=40-x.
\end{align}$
Area of the plot, $A$ , can be written as
$A=xy$.
Put $y=40-x$.
$\begin{align}
& A=x\left( 40-x \right) \\
& =-{{x}^{2}}+40x.
\end{align}$
And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-1,\ b=40.$
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:
$\begin{align}
& -\frac{b}{2a}=-\left( \frac{40}{-2} \right) \\
& =20.
\end{align}$
Put $x=20$ in equation for $y$.
$\begin{align}
& y=40-x \\
& =40-20 \\
& =20.
\end{align}$
Hence, for the area of the plot to be maximum the length of the plot would be $20$ yards and width would be $20$ yards.
The area of the plot would be
$\begin{align}
& A=xy \\
& =\left( 20\times 20 \right) \\
& =400y{{d}^{2}}
\end{align}$
