Q. 38 - Calculus

Question

A farmer wants to build four fenced enclosures on his farmland for his free-range ostriches. To keep costs down, he is always interested in enclosing as much area as possible with a given amount of fence. For the fencing projects given below,  determine how to set up each ostrich pen so that the maximum possible area is enclosed, and find this maximum area. 

      A rectangular ostrich pen is divided into six equal sections by two interior fences that run parallel to the east and west fences, and another interior fence running parallel to the north and south fences, as shown previously at the right. The farmer has allotted $2400$ feet of fencing material for this important project.

Step-by-Step Solution

Answer

Ans:   The maximum area is, $120,000\text{ square feet }$ 

Step 1. Given information.

given,  

      The rectangular ostrich pen is built with $2400$ feet of fencing material, divided into six equal sections by two interior fencings running parallel to the east-west fences another interior running parallel to the north-south fences.

Step 2. The objective is to determine the maximum possible area enclosed by the rectangular ostrich pen.

Let, the width be $x$ feet of each section and the length be $y$ feet of each section.

So its perimeter is,

     $\begin{array}{r}6\times 2(x+y)-y-y-x-x-x-y-y=2400\\ 12x+12y-4y-3x=2400\\ 9x+8y=2400\end{array}$

Solving the equation for $y$,  

   $\begin{array}{r}9x+8y=2400\\ 8y=2400-9x\\ y=300-\frac{9}{8}x\end{array}$

Now its area is,

    $\begin{array}{r}A=3x\cdot 2y\\ =3x\cdot 2\left(300-\frac{9}{8}x\right)\\ =3x\left(600-\frac{9}{4}x\right)\\ =1800x-\frac{27}{4}{x}^2\end{array}$

Step 3. To maximize the area find it's derivative first.

So,

   $\begin{array}{r}A=\frac{d}{dx}\left(1800x-\frac{27}{4}{x}^2\right)\\ A^{\mathrm{\prime }}=1800-\frac{27}{2}x\end{array}$

Equating to $0$,

    $\begin{array}{r}1800-\frac{27}{2}x=0\\ \frac{27}{2}x=1800\\ x=\frac{1800\times 2}{27}\\ x=133.33\end{array}$

Step 4. Putting x = 133.33  in  300 − 9 8 x

$\begin{array}{r}y=300-\frac{9}{8}\left(133.33\right)\\ y=150\end{array}$

The width of the pen is, 

    $\begin{array}{r}3\times 133.33\\ =399.99 \\ =400\text{ feet }\end{array}$

And its length is.

    $150\times 2=300 feet$

The maximum area is, 

    $\begin{array}{r}A=400\times 300\\ A=120,000\end{array}$

Therefore, the maximum area is $120,000$ square feet.