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Help with Minimum Area

  1. Mar 26, 2009 #1
    1) A cardboard box of 32in^3 volume with a square base and open top is to be constructed. Find the minimum area of cardboard needed.

    Since it has a square base, length and width must be the same, variable b for base.

    So volume could be written as: hb^2=32 and surface area as b^2+4bh=min.

    Once I get here, I''m not quite sure what to do.
    Any help is greatly appreciated!
     
  2. jcsd
  3. Mar 26, 2009 #2
    You want to get your minimum as a function of one variable. Use substitution to make that happen.
     
  4. Mar 26, 2009 #3
    Alright cool. I've got b^2 + 128/b^2 = min

    Should I take the derivative now?
     
  5. Mar 26, 2009 #4

    Dick

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    You might want to make sure that's correct before you take the derivative. You've got h*b^2=32 and you want to minimize b^2+4*b*h. Are you sure that means b^2+128/b^2 should be minimized?
     
  6. Mar 26, 2009 #5
    But aren't I supposed to substitute h into b2 + 4bh? and then solve for the critical point? Or would I just take the derivative of that? And then if thats what I'm supposed to do, do i take the derivative in relation to b or h?
     
  7. Mar 26, 2009 #6

    Dick

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    You are supposed to do exactly that thing. But are you sure substituting h=32/b^2 into b^2+4hb gives you b^2+128/b^2?
     
  8. Mar 26, 2009 #7
    Right, so b2 + 4b(32/b2)

    Which simplifies to

    b2 + 128b/b2

    Which simplifies to

    b2 + 128/b
     
  9. Mar 26, 2009 #8

    Dick

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    Right. Now take the derivative.
     
  10. Mar 26, 2009 #9
    so 2b - 128/b2

    and then would you multiply everything by b2 to clear the denominator? giving you 2b3 - 128 = 0
    So b = 3rt64

    So 4?

    Then if b = 4

    h must equal 2, per substitution into the original equation.

    So then the areas of the sides are 8, and the base is 16, so the total area is 8 x 4 + 16 = 48

    Is this right?
     
  11. Mar 26, 2009 #10

    Dick

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    It looks good to me. But I've made mistakes before. Never hurts to double check. Seem ok to you?
     
  12. Mar 26, 2009 #11
    Yeah, everything looks right. Thanks very much for your help!
     
  13. Mar 27, 2009 #12

    Mark44

    Staff: Mentor

    It would be a very good idea to work with equations, in part to help keep you focused on what you're doing.

    You're trying to minimize the area as a function of b, or A(b) = b2 + 4b(32/b2) = b2 + 128/b

    Then when you take the derivative, you get A'(b) = 2b - 128/b2

    To find the critical point, you're going to set A'(b) to 0, so 2b - 128/b2 = 0
    You can multiply both sides of that equation by b2 to get 2b3 - 128 = 0, and you can then solve that equation for b.
     
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