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Square-based pyramids surface area

  1. May 28, 2003 #1
    Among all square-based pyramids which have volume V = 1, which one has the smallest surface area?
     
  2. jcsd
  3. May 31, 2003 #2

    mathman

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    This is a messy elementary calculus problem. Define:
    a= length of base side
    h= height of pyramid
    t= altitude of triangle face
    from this t2=h2+(a/2)2

    Then

    V (volume) = a2h/3
    S (surface area) =a2+2at

    Set V=1, use the above formula for t, and let x=a2, we get:

    S=x+(36/x+x2)1/2

    S'=1+(x-18/x2)/(36/x+x2)1/2

    Set S'=0, we get x3=9/2

    To verify that this is a minimum (not maximum or horizontal inflection), observe that:
    x near 0, S approx 6/x1/2,
    x gets large, S approx 2x.

    Finally:
    S=(9/2)1/3+((9/2)2/3+36(2/9)1/3)1/2

    I suggest you work this through to understand how it goes.
     
  4. Jun 1, 2003 #3
    Wow, mathman. Thanks.
    I don't seem to understand this step:

    S=x+(36/x+x2)1/2
    S'=1+(x-18/x2)/(36/x+x2)1/2

    Shouldn't it read

    S'=1+(x-18/x2)/(36/x+x2)-1/2

    ...or something?
     
  5. Jun 1, 2003 #4

    mathman

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    You're right. I had the minus when I was doing it (pencil and paper), but I neglected to type it in. However, the final result still stands.
     
  6. Jun 1, 2003 #5

    Hurkyl

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    Er, isn't it right as is?

    dividing by z1/2 is the same as multiplying by z-1/2
     
  7. Jun 2, 2003 #6
    Oops! Yes. I overlooked the /.
     
  8. Jun 2, 2003 #7

    mathman

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    Hurkl got it right. When I read arcnets comment, I also forgot I had put in the /.
     
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