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1MileCrash

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## Homework Statement

Suppose that farmer brown has 260 feet of fencing and wants to build a pen that adjoins the whole side of his 100-ft barn (shown). What should the dimensions be for maximum area?

## Homework Equations

## The Attempt at a Solution

Alright, sorry no picture, but essentially we have a perimeter of 260, one of the sides must be at least 100 ft (adjoins the whole barn) and we want maximum area.

Well, my immediate thought it that a square always has the most area for any rectangle with a given perimeter, so I'm assuming we want what's AS CLOSE to a square as possible, which would be 100X30, yielding an area of 3000 square feet. But I suppose I need to prove this with calculus.

So.. any suggestions?

If perimeter is 260:

260 = 2x + 2y

y = 130 - x

Okay, y = 130 - x, and y must be at least 100. Y is greater than 100 when x is less than 30, so my answer is 30 because that's closest to a square, but let's prove it with calculus..

A = xy = x(130-x)

= 130x - x^2

A' = 130 - 2x

Okay, so now I need to find all critical points of the derivitive of area, where x =< 30, right?

so immediately, I have a critical point of 30, because my interval is (-infinity, 30]

Solve the derivative for 0,

0 = 130 - 2x

130 = 2x

65 = x

So we have a critical point of 65 (perfect square) and 30, but 65 is not in our interval, therefore the only critical point is 30, therefore 30 is the maximum within the interval, therefore the dimensions for maximum area are 100 x 30.

Is this adequate proof?

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