# Finding dimensions

• Mar 23rd 2008, 05:13 PM
suzdisp
Finding dimensions
I'm supposed to find the dimensions of a rectangular region with a maximum area that can be enclosed by 80 feet of fencing. this is what I have so far...

2x+ 2y=80
-2x -2x

2y=80-2x

y= 80-2x
------ --------> y= 40- (-1x)
2

Can somebody telling me if this is right so far?I got the -1x from the -2x
---
2

Is that correct so far? If not, could you please set me on the right track?
• Mar 23rd 2008, 05:14 PM
topsquark
Quote:

Originally Posted by suzdisp
I'm supposed to find the dimensions of a rectangular region with a maximum area that can be enclosed by 80 feet of fencing. this is what I have so far...

2x+ 2y=80
-2x -2x

2y=80-2x

y= 80-2x
------ --------> y= 40- (-1x)
2

Can somebody telling me if this is right so far?I got the -1x from the -2x
---
2

Is that correct so far? If not, could you please set me on the right track?

"y = 40 - (-1x)"
is not correct. (I presume it's just a typo.) It should be
$y = 40 - (1x)$
or
$y = 40 + (-1x)$

Do you know how to continue?

-Dan
• Mar 23rd 2008, 05:18 PM
suzdisp
well, I'll try. you tell me if I'm right or not

since you get y=40 -(1x)

you put that in for y, so: A=x(40-(1x))

A=40x-x^2?
• Mar 23rd 2008, 05:29 PM
topsquark
Quote:

Originally Posted by suzdisp
well, I'll try. you tell me if I'm right or not

since you get y=40 -(1x)

you put that in for y, so: A=x(40-(1x))

A=40x-x^2?

Yep. :)

Now what?

-Dan
• Mar 23rd 2008, 05:32 PM
suzdisp
well, I know you're supposed to find the derivatives and use it to determine the max. value of the area, but I have no idea how. I've started out by doing:

dA
---= 40x-(x^2)
dx

after that, I'm lost
• Mar 23rd 2008, 05:43 PM
topsquark
Quote:

Originally Posted by suzdisp
well, I know you're supposed to find the derivatives and use it to determine the max. value of the area, but I have no idea how. I've started out by doing:

dA
---= 40x-(x^2)
dx

after that, I'm lost

You have the right idea, but took your derivative wrong:
$A = 40x - x^2$

$\frac{dA}{dx} = 40 - 2x$

Now, what does the derivative of a function mean? It is the slope of the line tangent to the function at a given value of x. What happens to the derivative of a function when the function is at a relative maximum or minimum?

-Dan
• Mar 23rd 2008, 05:49 PM
suzdisp
I have no idea.
but I think the finally answer is x and y both = 20

that is what I have worked out. Am I right at all?
• Mar 23rd 2008, 05:52 PM
topsquark
Quote:

Originally Posted by suzdisp
I have no idea.
but I think the finally answer is x and y both = 20

that is what I have worked out. Am I right at all?

That is correct, but the question is "how did you find this out?"

When a function reaches either a relative max or min, the derivative vanishes. Thus
$40 - 2x = 0$

$x = 20$

Then you can find that y = 20 also.

-Dan
• Mar 23rd 2008, 06:01 PM
suzdisp
thats what I did to find 20. I found an example in my book.

then I solved for y. y=40-x

y=40-20
y=20

then, I put 20 in for y

20=40-1x
-40 -40

-20/ -1 =-1x/-1

20=x

YES!!!!!!!!!!!!!