Question

a graphing calculator is recommended. in this problem you are asked to find a function that models a real - life situation and then use the model to answer questions about the situation. a rancher with 950 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle (see the figure). (a) find a function a that models the total area of the four pens. (use w to represent the width of the field.) a(w)= (b) find the largest possible total area of the four pens. ft²

Explanation:

Step1: Set up the perimeter equation

Let the width of the field be $w$. Suppose the length of the entire rectangular - area is $l$. The total length of the fencing is used for 5 widths (since there are 4 internal dividers parallel to one side and 2 lengths). So, $5w + 2l=950$, and we can express $l=\frac{950 - 5w}{2}$.

Step2: Set up the area function

The area $A(w)$ of the rectangle (which is the total area of the four - pens) is given by the formula $A(w)=l\times w$. Substitute $l=\frac{950 - 5w}{2}$ into the area formula: $A(w)=w\times\frac{950 - 5w}{2}=\frac{950w-5w^{2}}{2}=475w-\frac{5}{2}w^{2}$.

Step3: Find the maximum of the area function

The function $A(w)=475w-\frac{5}{2}w^{2}$ is a quadratic function of the form $y = ax^{2}+bx + c$ where $a=-\frac{5}{2}$, $b = 475$, and $c = 0$. The vertex of a quadratic function $y = ax^{2}+bx + c$ has its $x$ - coordinate (in our case $w$ - coordinate) at $w=-\frac{b}{2a}$.
\[w=-\frac{475}{2\times(-\frac{5}{2})}=\frac{475}{5}=95\]

Step4: Calculate the maximum area

Substitute $w = 95$ into the area function $A(w)$.
\[A(95)=475\times95-\frac{5}{2}\times95^{2}\]
\[A(95)=475\times95-\frac{5}{2}\times9025\]
\[A(95)=45125-\frac{45125}{2}\]
\[A(95)=\frac{90250 - 45125}{2}=\frac{45125}{2}=22562.5\]

Answer:

(a) $A(w)=475w-\frac{5}{2}w^{2}$
(b) $22562.5$