Question

A rancher has 800 feet of fencing to put around a rectangular field and then subdivide the field into three identical smaller rectangular plot by placing to fence is parallel to the field shorter side. Find the dimensions that maximize the enclosed area. Write your answer as a fraction reduced to lowest term

Answer 1

The diagram of the problem is:

S is the length of the shorter side of the fence. L is the length of the longest side of the field.

We know that the perimeter of the rectangle is 800ft. This means:

$2S+2L=800$

And the area:

$A=SL$

The smaller rectangles will have dimensions:

The area is:

$a=\frac{SL}{3}$

As we can see, if we maximize the area of the bigger rectangle "A", we are also maximizing the area of the smaller rectangles "a".

Then, we have two equations:

$\begin{gathered} 2S+2L=800 \\ A=SL \end{gathered}$

We can solve for L in the first equation:

$\begin{gathered} 2S+2L=800 \\ 2L=800-2S \\ L=400-S \end{gathered}$

Then substitute in the second:

$A=S(400-S)$

Simplify:

$A=400S-S^2$

This is a function of the area depending on the length of the shorter side of the rectangle:

$A(S)=400S-S^2$

We can find the maximum of this function if we find the value where the derivative of this function is 0.

Let's differentiate:

$A^{\prime}(S)=400-2S$

And now we find where A'(S) = 0:

$\begin{gathered} 0=400-2S \\ 2S=400 \\ S=200 \end{gathered}$

We have found that the shorter side must have a length of 200ft to maximize the area. Let's find the length of the larger side:

$L=400-200=200$

As expected, the quadrilateral which maximizes the area is the square. Thus, the dimensions of the field are 200ft x 200ft