Optimization

[Caldus]

If the surface area equation for a box is given as 7200 = 6x^2 + 4xy where x represents the length and width of the box while y represents the height, then what is the maximum volume that can be calculated?

My answer was 540000, but not sure.

[Inspector Gadget]

Did you come up with the equation on your own?

Because I would think that, with length and width being x and y being height, the surface area would be 7200 = 4xy + 2x^2.

Anyway, with what you gave, I found the max. volume to be something else...

Volume: V = x^2 * y
Surface Area: 7200 = 6x^2 + 4xy

Solve surface area for y: (7200 - 6x^2)/4x, simplify it a bit, (3600 - 3x^2)/2x

Substitute it in to the volume...

V = x^2 * (3600 - 3x^2)/2x
V = [x(3600 - 3x^2)]/2
V = (3600x - 3x^3)/2
V = 1800x - (3/2)x^3

Take the derivative...
V' = 1800 - (9/2)x^2

Set equal to 0 and solve for x...
1800 - (9/2)x^2
1800 = (9/2)x^2
3600 = 9x^2
400 = x^2
x = +/- 20

And, because we can't have a negative length, x = 20.

Next, find y by going back into the surface area equation...

7200 = 6x^2 + 4xy
7200 = 6(20)^2 + 4(20)y
7200 = 2400 + 80y
4800 = 80y
60 = y

Now, back into the volume...
V = x^2 * y
V = (20)^2 * 60
V = 400 * 60
V = 24000

Someone may want to double check that as I always make some mistake to screw up my work.

[Ebolamonk3y]

I got 24,000...

[Parth Dave]

nope, its fine, i got the same thing

[Ebolamonk3y]

How did you get 540,000?

[Caldus]

Oh my God, I realized that I made a stupid little mistake. I will show you where I messed up:

OK, let me give the actual details of the problem. Maybe I misinterpreted the problem itself.

You have $7200. It will cost you $1 per square meter to cover the bottom part of a box. It will cost you $5 per square meter to cover the top part of a box. For every other side, it will cost you $1 per square meter to cover it. Give the maximum possible volume for this box with the amount of money you have.

So my equation ends up being: 7200 = 4xy + x^2 + 5x^2 which simplifies to 7200 = 4xy + 6x^2
Then I calculated the area to be x^2y.

In the surface area equation, I solve for y, so area ends up equaling:

x^2(((7200 - 6x^2)/4x))

I simplified to:

(7200x^2 - 6x^4)/4x
1800x - (3/2)x^3

Then I got the derivitive of area:

1800 - (9/2)x^2

Set it to 0 and find the max value:

0 = 1800 - (9/2)x^2
-1800 = -(9/2)x^2 <--- Right here I accidently interpreted the coefficient to be 3 because for some reason I was thinking (6/2)x^2 instead of (9/2)x^2.
x = 20

Plug 20 back in the original area equation with only x's and you get 24,000.

And I apparently I got the same answer as you guys now that I realize my mistake.

Uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu uuuuuuuuuuuuuuuuuggggggggggggggggggggggggggggggggg ggggggggggggggggggggggggggggggggggggggggg! Why did I have to do this on my test! I know this stuff! Gaaaaaaaaah!

Sorry. ; )

[HallsofIvy]

Glad you got the right answer.

By the way, in your last response you were consistently saying "area" when you meant "volume".

[Caldus]

Yep, you're right.

I have always been very good about making little mistakes like the one you just mentioned as well as the mistake I made in the problem. My test grades would be so much better otherwise, LOL.