1. ## Pratical 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?

2. Originally Posted by jigger1
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?
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?

3. Thanks Jameson. That is were I kind of get stuck...

4. Originally Posted by jigger1
Thanks Jameson. That is were I kind of get stuck...
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.

5. 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!

6. Originally Posted by jigger1
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!
For 2x+y = 600

Similar idea holds:

$\displaystyle (\sqrt{2x}-\sqrt{y})^2 \geq 0 \Rightarrow \sqrt{2A} \leq 300 \Rightarrow A \leq 45000$
Equality holds when 2x=y.

Thus maximum area is 45000 and it occurs when x = 150, y =300

7. 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

8. Got it, thanks for all your help. You guys (girls) ROCK!