Challenging Math question

Question

A farmer has 1600 feet of fence and wants to enclose a rectangular field with two internal fences parallel to one side. What are the dimensions that would maximize the area?

Answer 1

Ishan,

First you need to establish equations that relate the length of fence to the area.

 

A = W * L   gives us the entire enclosed area. We don't care about the subareas.

 

The total length of fence includes not just the perimeter, but also the two dividers which are the same as the width of the rectangle. So using F for the total length of fence, we get:

 

F = 4W + 2L   or   1600 = 4W + 2L

 

Any function which has a defined maximum will have a peak in its graph at that point. But the area equation currently has three variables, and thus cannot be graphed. So we need to substitute from the fence equation into the area equation so that area is defined in terms of only one other variable.

 

1600 - 2L = 4W            solve for W

 

400 - .5L = W              now substitute

 

A = L(400 - .5L)          distribute

 

A = -.5L² + 400L         We see here that the graph is a parabola that opens down. Thus we know that the peak we are looking for will be found at the vertex of the parabola. We can find the x coordinate (L) of the vertex  with the formula:

 

x_v = -b/(2a) = -400/(2(-.5)) = 400 = L  Now plug into the fence equation.

 

W = 400 - .5(400) = 200

 

Final dimensions:  200' x 400' giving the max area as 80,000 ft²

 

I hope that helps.