Can you please help me with this question? I'm not sure I get the picture here.

Question

A farmer has 200 feet of fencing to surround a small plot of land. He wants to maximize the amount of space possible using a rectangular formation. The enclosed area will be against the side of a barn so he only needs one of the lengths of the rectangular area to be enclosed by fencing. What is the maximum area that can be enclosed by the fence?

Answer 1

Hi Sona,

This is a tough problem that actually requires calculus to solve.

Assuming you don't know calculus, then you need to know that the shape of the maximum area of a rectangle is a square. Another way of saying that, is the width of the shape that produces the maximum area of a rectangle is equal to one fourth of the perimeter.

The problem says that one of the lengths is the side of the barn so no fencing is required on that side. Therefore you will have two widths and one length of fencing that add up to 200 feet.

One fourth of the perimeter = 50

The other width also equals 50. So the length of fencing = 200 - 50 - 50 = 100

So the area = 50 x 100 = 5000 square feet.

To convince yourself that this is the correct answer, you can add or subtract one foot from the width and come up with the appropriate length and multiply them and you'll find you get an area that is slightly less than 5000.

Answer 2

A picture could look like this:

The perimeter of the fence must be < 200 feet because that's all he has and it makes sense that if he wants to maximize the area, he would want to use all of the fencing available, so P = 200. P also equals 2W + L so 2W + L = 200

Since area is defined as L•W and we want to maximize that area, then we will want to consider the graph of the area as a function of either length or width and find the largest area available considering the constraint of a 200 foot perimeter.

So, since 2W + L = 200, we can say L = 200 - 2W and we can plug "200 - 2W" in for "L" in our A = L•W equation.

A = L•W

A = (200 - 2W)W

A = -2W² + 200W

This is a quadratic equation that graphs as an inverted parabola (with a maximum at the vertex). So we want to find that maximum. The vertex of a parabola is found using -b/2a so W_VERTEX = -200/[(2)(-2)] = 50

If W = 50, then we can plug that into L = 200 - 2W to find L

L = 200 - 2(50) = 100

So the dimensions are 50 x 100 and that yields 5000 sq ft.

Just for grins, let's pick a width of 40 feet just to see what the area would be. L = 200 - 2(40) = 120 so the area would be 40•120 = 4800 sq ft. This supports our solution that the max is 5000 sq ft.