College Algebra, Chapter 4, 4.1, Section 4.1, Problem 72

A rancher with $750$ ft of fencing wants to enclose a rectangular area and divide into four pens with fencing parallel to one side of the rectangle.







a.) Find a function that models the total area of the four pens.

b.) Find the largest possible area of the four pens.

a.) If the area of one pen is $A = xy$, then the total area of the four pens is $A_T = 4xy$. Since the $750$ ft of fencing material corresponds to the perimeter of the lot, then,


$\begin{equation} \begin{aligned} P =& 8x + 5y \\ \\ 750 =& 8x + 5y \end{aligned} \end{equation}$


Solving for $y$

$\displaystyle y = \frac{750 - 8x}{5}$

Thus,


$\begin{equation} \begin{aligned} A_T =& 4xy = 4x \left( \frac{750 - 8x}{5} \right) = 600x - \frac{32}{5} x^2 \\ \\ A_T =& 600x - \frac{32}{5} x^2 \end{aligned} \end{equation}$


b.) The function $A_T$ is a quadratic function with $\displaystyle a = - \frac{32}{5}$ and $b = 600$. Thus, its maximum value occurs when

$\displaystyle x = \frac{-b}{2a} = \frac{-600}{\displaystyle 2 \left( \frac{-32}{5} \right)} = \frac{375}{8}$ ft

Therefore, $A_T$ is maximum at..


$\begin{equation} \begin{aligned} A_T = 600x - \frac{32}{5} x^2 =& 600 \left( \frac{375}{8} \right) - \frac{32}{5} \left( \frac{375}{8} \right)^2 \\ \\ =& \frac{28125}{2} \text{ ft}^2 \end{aligned} \end{equation}$