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Maximizing the volume of a square Pyramid?

  1. Jan 18, 2012 #1
    1. The problem statement, all variables and given/known data

    I am given an 8cm by 8cm piece of paper. I have to cut out a square-based pyramid out of that that gives me the greatest volume.

    2. Relevant equations

    I know that the volume is V = 1/3 * b^2 * h
    b = base
    h = height

    I know that the surface area is A = b^2 + 2bh

    3. The attempt at a solution

    I slightly remember optimization. I only remember you need to combine the two but that's it?
  2. jcsd
  3. Jan 18, 2012 #2


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    The surface area isn't involved in this solution. Of course I realize that the surface area of the pyramid can't be greater than 64 cm2.

    Connect the four midpoints of the edges of the sheet of paper. Let the resulting square be the base of a pyramid. Fold the corner triangles upward & inward until the corners of the original 8×8 piece of paper meet at the apex of a pyramid. That square pyramid is one with maximum base area. You would hardly call it a pyramid at all, because its altitude is zero.

    Other square pyramids will require that you use a smaller base, but will result in larger altitude. Such pyramids will have surface areas less than 64 cm2.
  4. Jan 22, 2012 #3
    yeah i experimented and found out that the smaller you cut from the square, the larger the volume(aka larger base = larger volume). Larger altitudes give you smaller volumes.

    How would I find out the maximum volume using calculus work though?

  5. Jan 22, 2012 #4


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    Come up with a formula relating the volume to some parameter related to how much is cut out, or related to the length of an edge of the square base, or related to the area of the base, or related to ...

    So the volume will be a function of whatever parameter you choose.

    Then use calculus to find the maximum value of that function.
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