I need to create a fenced area along the side of a barn. Six hundred feet of fencing is available to use.

What are the dimensions which give the maximum area?

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- Apr 28th 2008, 01:17 PM #1

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- Apr 28th 2008, 01:22 PM #2

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You have two variables here so we're going to use two equations, substitute, and then find the maximum. Start with area. If the width of the fence is x and the length is y, than $\displaystyle Area=xy$. This is what you trying to maximize. Now let's talk about the perimeter. We have 600 ft. of fencing to use, thus we know that $\displaystyle 2x+2y=600$. From here, solve for a variable from the perimeter equation and substitute that into the area one so that the area equation has just one variable. Then take the derivative and you know the rest.

EDIT: Ooops. This is in the pre-calc forum. Do you know calculus?

- Apr 28th 2008, 02:29 PM #3

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- Apr 29th 2008, 02:21 AM #4
We know that

$\displaystyle x+y=300$

$\displaystyle A = xy$

Without calculus:

$\displaystyle (\sqrt{x}-\sqrt{y})^2 \geq 0 \text{ [Equality at x = y] }$

$\displaystyle \Rightarrow x+y \geq 2\sqrt{xy} $

$\displaystyle \Rightarrow 300 \geq 2\sqrt{A} $

$\displaystyle \Rightarrow \sqrt{A} \leq 150 $

$\displaystyle \Rightarrow A \leq 25500 \text{ [Equality at x = y] }$

$\displaystyle \text{ [Equality at x = y] and }x+y = 300 \Rightarrow x=y=150$

So maximum area is 25500 square units and that occurs for a square fencing of side 150 units.

- Apr 30th 2008, 03:40 PM #5

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- May 1st 2008, 02:15 AM #6

- May 1st 2008, 03:27 AM #7

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2x + y = 600 (1)

also

Area = xy

sub in from 1

Area = x(600-2x)

Area = -2x^2 + 600x

complete the square gives

Area = -2(x-150)^2 + 45 000

so the vertex of the parabola will be (150, 45 000)

going back to this equation: 2x + y = 600

we know the x value for the max area is 150

2(150) + y = 600

y = 600 - 300

y = 300

we can check to confirm using the area equation A = xy

45 000 = 150(300)

45 000 = 45 000

check

- May 1st 2008, 02:15 PM #8

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