Answer
Length of plot $=100ft$
Width of plot $=66.66ft$
Maximum area of plot $=6666f{{t}^{2}}$
Work Step by Step
Let us assume the length of the plot be $y$ and breadth of the plot along the river be $x$.
The plot is to be fenced from all sides and is divided into two by another fence parallel to one side. The total available length of fence is $600ft$.
$3x+2y=400$.
And solve for $y$.
$\begin{align}
& 2y=400-3x \\
& y=200-1.5x.
\end{align}$
Area of the plot, $A$, can be written as
$A=xy$.
Put $y=200-1.5x$
$\begin{align}
& A=x\left( 200-1.5x \right) \\
& =-1.5{{x}^{2}}+200x.
\end{align}$
And compare with the quadratic function, $f\left( x \right)=a{{x}^{2}}+bx+c$. It gives $a=-1.5,\ b=200$.
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:
$\begin{align}
& -\frac{b}{2a}=-\left( \frac{200}{-3} \right) \\
& =66.66
\end{align}$
Put $x=66.66$ in equation for $y$.
$\begin{align}
& y=200-1.5x \\
& =200-1.5\left( 66.66 \right) \\
& =100
\end{align}$
Hence, for the area of the plot to be maximum the length of the plot would be $100ft$ and width would be $66.66ft$.
The area of the plot would be
$\begin{align}
& A=xy \\
& =\left( 100\times 66.66 \right)f{{t}^{2}} \\
& =6666f{{t}^{2}}
\end{align}$
