What is the maximum area possible? Explain.?

Question

You have a single 440 ft roll of fencing available to fence in a rectangular storage area. Your boss would like the area to be as large as possible, so yoou plan to use all the fencing.

Width: 50 100 150
Length: 170 150 70
Area: 8500 12000 10500

A(w)= w(220-w)

What is the maximum area possible? Explain.

Answer 1

perimeter = 2w + 2l = 440
w + l = 220

A = lw = w(220-w) = 220w - w^2
dA/dw= 220-2w
maximum point occurs at dA/dw=0
2w=220
w=110

Maximum area= 110(220-110) = 12100sq. ft

I'm not quite sure whts the string of numbers about

Answer 2

Assuming that your boss doesn't want a very large area surrounded by a very small fence there is no point in cutting your roll up.

What on earth is all this width, length,area stuff?

Is the region in which you are to put your fence of vast extent and flat?

If so make a circular fence.

What has A(w) = w(220-w) to do with anything? Ah, I see, it's finding the area of a rectangle of width w and periphery 440 and it's maximised when w=110 - but that's just for rectangles, circles are better. The area of a circle of circumference =440 is (440/pi)^2/4*pi = 220^2/pi .

Answer 3

Maximum area possible for a rectangle shape would be a square. The dimensions for a 440 ft roll of fencing would be 110x110. That would yield 12100 square ft. You will notice the more you deviate from a square for the less area you will have. 100 x 120 12000 square ft 50 x 170 = 8500 square ft.
Of course if you remove your requirement of a rectangle then the optimal shape is a circle. The closer you get to a circle the more area you will have.

Answer 4

A(w) = 220w - w^2

dA/dw = 220 - 2w

Let dA/dw =0

=> 220 - 2w = 0
2w = 220
w = 110

Therefore width and length has to be 110 to maximize area.

Area = 110 * 110 = 12100

Answer 5

p = 2w + 2l = 440
w + l = 220

A = lw = w(220-w) = 220w - w^2
dA/dw = 220-2w = 0
w = 110
l = 110

A = wl = 110^2 = 12100 sq. ft