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Maximum volume that can be calculated?

  1. May 3, 2004 #1
    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. jcsd
  3. May 3, 2004 #2
    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.
  4. May 3, 2004 #3
    I got 24,000...
  5. May 3, 2004 #4
    nope, its fine, i got the same thing
  6. May 3, 2004 #5
    How did you get 540,000?
  7. May 4, 2004 #6
    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. ; )
  8. May 4, 2004 #7


    User Avatar
    Science Advisor

    Glad you got the right answer.

    By the way, in your last response you were consistently saying "area" when you meant "volume".
  9. May 4, 2004 #8
    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.
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