# Maximum volume that can be calculated?

1. May 3, 2004

### 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.

2. May 3, 2004

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.

3. May 3, 2004

### Ebolamonk3y

I got 24,000...

4. May 3, 2004

### Parth Dave

nope, its fine, i got the same thing

5. May 3, 2004

### Ebolamonk3y

How did you get 540,000?

6. May 4, 2004

### 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.

Uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuugggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggggg! Why did I have to do this on my test! I know this stuff! Gaaaaaaaaah!

Sorry. ; )

7. May 4, 2004