Q. 41 - Calculus

Question

You are in charge of constructing a zoo habitat for prairie dogs, with the requirement that the habitat must enclose $2500$ square feet of area and use as little border fencing as possible. For each of the habitat designs described below, find the amount of border fencing that the project will require.

             A trapezoid-shaped habitat whose angled side is an enclosed walkway for zoo patrons (so that no fencing is needed along the walkway), where the walkway makes an angle of $60°$ with the right fence, as shown next at the left.

Step-by-Step Solution

Answer

Ans:   The length of the fencing is approximately $143$ feet

Step 1. Given information.

given,  

       The maximum area is $2500$ square feet.  

Step 2. Consider the trapezoid with a non-parallel side measuring y feet and a smaller parallel side measuring x feet as shown above.

The length of the other non-parallel side is  and the longer parallel side is,

    $x+y \cos {60}^{\circ }$

The objective is to minimize the sum of the three sides of the trapezoid.

Step 3. The area of the trapezoid is,

$\begin{array}{r}A=\frac{1}{2}\left(x+x+y\cos {60}^{\circ }\right)\left(y\sin {60}^{\circ }\right)\\ =\frac{1}{2}\left(2x+\frac{y}{2}\right)\left(\frac{\sqrt{3}}{2}y\right)\\ =\frac{\sqrt{3}y(4x+y)}{8}\end{array}$

The area of the trapezoid should be $2500$ square feet.

Therefore,

       $\begin{array}{r}\frac{\sqrt{3}y(4x+y)}{8}=2500\\ \sqrt{3}y(4x+y)=20000\end{array}$

Step 4. The sum of three sides other than the wall is,

    $\begin{array}{r}S=x+y\sin {60}^{\circ }+x+y\cos {60}^{\circ }\\ =2x+\frac{y}{2}(\sqrt{3}+1)\end{array}$

Substitute for $x$ in terms of $y$ from the area relation,

      $\begin{array}{r}S\left(y\right)=2\left(\frac{5000}{\sqrt{3}y}-\frac{y}{4}\right)+\frac{y}{2}(\sqrt{3}+1)\\ =\frac{10000}{\sqrt{3}y}-\frac{y}{8}+\frac{y}{2}(\sqrt{3}+1)\end{array}$

Find the value of x ,   y as follows,

Differentiate $S\left(y\right)$ with respect to $y$,

     $S^{\mathrm{\prime }}\left(y\right)=-\frac{10000}{\sqrt{3}{y}^2}-\frac{1}{8}+\frac{\sqrt{3}+1}{2}$

Equate the derivate to zero and solve for $y$,

     $\begin{array}{r}-\frac{10000}{\sqrt{3}{y}^2}-\frac{1}{8}+\frac{\sqrt{3}+1}{2}=0\\ y^2=\frac{10000}{\sqrt{3}\left(\frac{\sqrt{3}+1}{2}-\frac{1}{8}\right)}\\ y\approx 68.2\end{array}$

Therefore, the value of $x$ is,

      $\begin{array}{r}x=\frac{5000}{\sqrt{3}\left(68.2\right)}-\frac{68.2}{4}\\ \approx 25.3\end{array}$

Step 6. Substitute the value of x and y in S to get the length of the fence,

$\begin{array}{r}S=2\left(25.3\right)+\frac{68.2}{2}(\sqrt{3}+1)\\ \approx 143\end{array}$

Therefore, the length of the fencing is approximately $143$ feet