Q. 42 - 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. 

      An arena-style trapezoid-shaped habitat whose long backside is a wall with a landscape mural (so no fencing is needed along the back wall), where the back wall makes an angle of $60°$ with the slanted side fences, as shown previously at the right.

Step-by-Step Solution

Answer

Ans:  The length of the fencing is approximately $186$ 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 distance between parallel sides is $y\sin {60}^{\circ }$ and the longest parallel side is,

$x+2y\cos {60}^{\circ }$

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

Step 3. The area of the trapezoid is,

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

Consider the area of the trapezoid is $2500$ square feet.

Therefore,

   

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

   $\begin{array}{r}S=y+x+y\\ =x+2y\end{array}$

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

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

Step 5. Find the value of x ,   y as follows,

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

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

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

     $\begin{array}{r}-\frac{10000}{\sqrt{3}{y}^2}+\frac{3}{2}=0\\ y^2=\frac{20000}{3\sqrt{3}}\\ y\approx 62\end{array}$

Therefore, the value of $x$ is,

 $\begin{array}{r}x=\frac{10000}{\sqrt{3}\left(62\right)}-\frac{62}{2}\\ \approx 62\end{array}$

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

$\begin{array}{r}S=62+2\left(62\right)\\ =186\end{array}$

Therefore, the length of the fencing is approximately $186$ feet.