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Maximizing volume of a Square-Based Pyramid?

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data
    I am given an 8in x 8 in piece of paper and I need to build a pyramid with a square base with the greatest volume.




    2. Relevant equations
    V = 1/3 * x^2 * h

    b = base length
    h = height length

    I'm trying to get everything in terms of x, which is the length of the base of the square I need to make.

    The slant length is l

    w(length of the entire slant) = 2l + x


    http://i39.tinypic.com/2me4jsw.jpg

    Oh and l = h^2 + x^2/2. height squared + one half x^2.



    3. The attempt at a solution

    This is an optimization problem, so I put everything in terms of one variable, and then get the derivative



    I just don't know if this is the derivative.

    http://i44.tinypic.com/2rh07f5.jpg
     
  2. jcsd
  3. Jan 30, 2012 #2

    ehild

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    Hi Sydneygte,


    You made a little mistake: It should be (x/2)^2 instead of x^2/2.
    Also, type in the expression of the volume in terms of x and its derivative. You picture is not clear.

    ehild
     
  4. Jan 30, 2012 #3
    ohh that changes it a bit

    so h^2 = l^2 - (x/2)^2

    so you can square root everything so

    h = l - (x/2)? that doesn't seem right
     
  5. Jan 30, 2012 #4

    SammyS

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    If square roots worked like that, Pythagoras would have been out of business long ago.

    Square roots don't work like that.

    After all,
    9 = 25 - 16 .​
    But:
    3 ≠ 5 - 4 .​
     
  6. Jan 30, 2012 #5
    so what would the equation look like if you were trying to get H to the power of one?

    h = sqrt(l^2 - .5x^2)?
     
  7. Jan 30, 2012 #6

    SammyS

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    It would be:
    h = sqrt( l2 - (0.5x)2 )
    or equivalently
    h = sqrt( l2 - (0.25)x2 )
    (That is a lower case L, not a number 1.)
     
  8. Jan 30, 2012 #7
    so then the function in terms of x would be

    V = 1/3X^2 * sqrt( sqrt(128)-x)^2 - (0.5x)^2) ?

    that seems confusing haha. has would I simplify the double square root? Since it is being squared in the inside can I make it into

    V = 1/3X^2 * sqrt(128 - X^2 - 0.25X^2)?
     
  9. Jan 30, 2012 #8

    SammyS

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    You can't do that either .

    (a-b)2 a2 - b2 .

    But the following is true:

    (a-b)2 = a2 -2ab + b2

    You will still have a square root if you "FOIL" ( sqrt(128)-x)2.

    BTW, √(128) = (8)√(2), and either way you write this, it's just a number, there's no variable involved.

    Also , you are missing a parenthesis in " sqrt( sqrt(128)-x)^2 - (0.5x)^2) ".

    It should be: sqrt( ( sqrt(128)-x)^2 - (0.5x)^2) .
     
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