## Precalculus (6th Edition) Blitzer

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