## Precalculus (6th Edition) Blitzer

The dimensions of the rectangle that maximize the area of the rectangular portion of the field are $110$ yards by $\frac{220}{\pi }$ yards.
Consider that the perimeter of the track is $440$ yards. Let the length of the rectangle part be x. And let the width of the rectangle part be y. Thus, the diameter of the semicircle is y. Recall that the perimeter of the rectangle is twice the sum of its length and width and the perimeter of the circle is twice the product of its radius and $\pi$. Thus, $440=2x+\pi y$. Solve for y, $y=\frac{440-2x}{\pi }$ Thus, the width of the rectangle is $y=\frac{440-2x}{\pi }$. Recall that the area A of the rectangle is the product of length and width. Thus, \begin{align} & A=x\left( \frac{440-2x}{\pi } \right) \\ & =\frac{-2}{\pi }{{x}^{2}}+\frac{440}{\pi }x \end{align} Maximize, $A\left( x \right)=\frac{-2}{\pi }{{x}^{2}}+\frac{440}{\pi }x$ Compare it with the standard form $f\left( x \right)=a{{x}^{2}}+bx+c$ . The value of $a$ is $\frac{-2}{\pi }$ , $b$ is $\frac{440}{\pi }$ and $c$ is $21000$ . Since, $a<0$ , the function has a maximum value at $x=\frac{-b}{2a}$. \begin{align} & x=\frac{-b}{2a} \\ & =\frac{-\left( \frac{440}{\pi } \right)}{2\left( \frac{-2}{\pi } \right)} \\ & =\frac{440}{4} \\ & =110 \end{align} Substitute $110$ for x in $y=\frac{440-2x}{\pi }$ . \begin{align} & y=\frac{440-2x}{\pi } \\ & =\frac{440-2\left( 110 \right)}{\pi } \\ & =\frac{440-220}{\pi } \\ & =\frac{220}{\pi } \end{align} Hence, the dimensions of the rectangle that maximize the area of the rectangular portion of the field are $110$ yards by $\frac{220}{\pi }$ yards.