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Math Help - Minimum Value - Lidless box with square ends

  1. #1
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    Minimum Value - Lidless box with square ends

    A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area of the metal used for which the volume of the box is 3.5m^3.

    This is my solution:
    A = x^2 + 4xh
    Substitute h = 3.5/x^2 into A
    So A = x^2 + 14x^-1
    dA/dx = 2x - 14x^-2
    Equating dA/dx = 0, so x = 1.91cm
    Therefore, the least area of the metal used = 10.98cm^2
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  2. #2
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    Quote Originally Posted by darkvelvet View Post
    A lidless box with square ends is to be made from a thin sheet of metal. Determine the least area of the metal used for which the volume of the box is 3.5m^3.

    This is my solution:
    A = x^2 + 4xh
    Substitute h = 3.5/x^2 into A
    So A = x^2 + 14x^-1
    dA/dx = 2x - 14x^-2
    Equating dA/dx = 0, so x = 1.91cm
    Therefore, the least area of the metal used = 10.98cm^2
    You have assumed a square base but the question says square ends ....
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  3. #3
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    Also, a couple of comments- never write equations without saying what you are taking the symbols to mean:
    If you had specifically written "x= length of base, h= height" you might have caught your mistake.

    Before writing h= 3.5/x^2, you should first state where you got that from: V= w*b*h and you were given that the volume is 3.5 m^3.

    Finally, I notice that you put units for each measurement- Good! But pay attention to the units. If the volume was given in cubic meters, m^3, are you sure the length of a side should be in centimeters?
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  4. #4
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    Thanks so much. Was really careless with the square ends bit. Thanks for the advice.
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