## Precalculus (6th Edition) Blitzer

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