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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- April 28th 2008, 01:17 PMjigger1Pratical Problem
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? - April 28th 2008, 01:22 PMJameson
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 . 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 . 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? - April 28th 2008, 02:29 PMjigger1
Thanks Jameson. That is were I kind of get stuck...

- April 29th 2008, 02:21 AMIsomorphism
- April 30th 2008, 03:40 PMjigger1
How would I solve w/ calculas?

Also, for perimeter would that not be 2x+y=600 because I will be fencing along side of a barn?

Thanks! - May 1st 2008, 02:15 AMIsomorphism
- May 1st 2008, 03:27 AMfinch41
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 PMjigger1
Got it, thanks for all your help. You guys (girls) ROCK!