A farmer wishes to put a fence around a rectangular field and then divide the field?

Let w be the entire width, h be the entire height, and (w/3) be one of the three

sections of the width. The width is divided in thirds so there must be 4 heights

of fence.

Then 4h + 2w = 1000 .... Divide both sides by 2

2h + w = 500

w = 500 - 2h

Now A = 3 * (w/3) * h = 3 * ((500 - 2h) / 3) * h = 500h - 2h^2

dA/dh = 500 - 4h .... Set this = 0.

500 - 4h = 0

h = 125

So the height (vertical) is 125

Since w = 500 - 2h, w = 250

There are two horizontal widths totaling 500 feet and four heights totaling 500 feet.

So the fence used is exactly 1000 feet

and the area is 250 * 125 = 31,250 sq. ft.

Since dA/dh = 500 - 4h, d^2A/dh^2 = -4 which is negative so this is a maximum.

.

So there are 6 equal pieces of fence going one way, but only 2 going the other. Call each of the 6 pieces x; that's 6x total. This leaves 3000 - 6x to split between the other two sides so each of them is (3000 - 6x)/2 = 1500 - 3x. The area is now x(1500 - 3x) = 1500x - 3x^2. The maximum is found on the axis of symmetry, whose equation is x = -b over 2a Here, that's x = -1500 / 2(-3) = -1500/-6 = Go from here

A = LW or L = A/W

1000 = 2L + 4W

500 = L + 2W

500 = A/W + 2W

A = 500W - 2W^2

A' = 500 - 4W & A'' = -4, so set A' to 0 for maxima

0 = 500 - 4W, W = 125, L = 500-2W = 250

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L = 250 yds

W = 125 yds

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