# College Algebra

A rectangular enclosure is to be constructed with 2,000 feet of fence in order to have four sections, as in the diagram below. What should be the dimensions in order to maximize the area enclosed? What is the maximum area? (So the diagram is just four rectangles connected side by side.
Sort of like | | | | | but with tops and bottoms too).

I totally drew a blank on this I'm sorry for not showing my work I don't know where to start.

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1. well, if each pen is x by y, then the vertical lines have length x, and the horizontal lines have length 4y, then we know

5x+8y = 2000
then the area is

a = 4xy = 4x(2000-5x)/8 = 1000x - 5/2 x^2

This just a parabola, and its vertex will be the maximum area. Work it out, and you will see that as in all problems like this, the area is maximum when the fencing is divided equally among lengths and widths.

In this case, that means 1000 ft for the 5 vertical fences (x=200) and 1000 ft for the horizontal fences (y=125).

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posted by Steve
2. So the maximum area is the vertex? Is the vertex 200, 1000?

And the dimensions in order to maximize the area enclosed are 200, 125?

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posted by Sasha
3. well of course the vertex is the maximum area. The parabola represents the area, right?

And you are correct in your final answers, with the exception that when x=200 and y=125, 4xy = 4*200*125 = 100,000

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posted by Steve

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