Q. 37 - 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 built with $1000$ feet of fencing material, divided into three equal sections by two interior fences that run parallel to the exterior side fences, as shown next at the left.

Step-by-Step Solution

Answer

Ans:   The maximum area is $31,250$ square feet. 

Step 1. Given information.

given,   

        The rectangular ostrich pen is built with $1000$ feet of fencing material, divided into three equal sections by two interior fencings running parallel to the exterior side 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,

$\begin{array}{r}2(x+y)+2(x+y)-y+2(x+y)-y=1000\\ 2x+2y+2x+2y-y+2x+2y-y=1000\\ 6x+4y=1000\\ 3x+2y=500\end{array}$

Solving the equation for $y$, 

      $\begin{array}{r}3x+2y=500\\ 2y=500-3x\\ y=\frac{500-3x}{2}\end{array}$

Now, its area is, 

      $\begin{array}{r}A=3x\cdot y\\ =3x\left(\frac{500-3x}{2}\right)\\ =\frac{1500x-9x^2}{2}\end{array}$

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

So,

     $\begin{array}{r}{A}^{\mathrm{\prime }}=\frac{d}{dx}\left(\frac{1500x-9x^2}{2}\right)\\ A^{\mathrm{\prime }}=750-9x\end{array}$

Equating to $0$,

  $\begin{array}{r}750-9x=0\\ 9x=750\\ x=83.33\end{array}$     

Step 4. Putting x = 83.33  in  y = 500 − 3 x 2 ,

$\begin{array}{r}y=\frac{500-3\left(83.33\right)}{2}\\ y=125.005\\ y=125\end{array}$

The width of the pen is,

    $\begin{array}{r}3\times 83.33 \\ =249.99 \\ =250\text{ feet }\end{array}$

And its length is $125$ feet

The maximum area is,

   $\begin{array}{r}A=125\times 250\\ A=31250\end{array}$

Therefore, the maximum area is $31,250$ square feet.