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Math Help - Optimization problem

  1. #1
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    Optimization problem

    A closed box with a square base is designed to have a surface area of 216 sq. ft. Find the dimensions of the box if the volume is to be a max.

    Pardon my elementary approach to this.. but this is how I have been doing these sort of problems.. i make up 2 equations ...

    6lw = 216 sq ft (surface area)
    and
    l^3 = maximum (volume of a cube, square on all sides)

    then i solve for one of the variables, in this case L (length)
    so 216/6w = 36/w = l

    then i substitute

    (36/w)^3 = maximum

    do you guys follow? in this particular case, i get some weird numbers in the end, so i think my approach didn't work in this problem?
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  2. #2
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    Quote Originally Posted by alny View Post
    A closed box with a square base is designed to have a surface area of 216 sq. ft. Find the dimensions of the box if the volume is to be a max.

    Pardon my elementary approach to this.. but this is how I have been doing these sort of problems.. i make up 2 equations ...

    6lw = 216 sq ft (surface area)
    and
    l^3 = maximum (volume of a cube, square on all sides)

    then i solve for one of the variables, in this case L (length)
    so 216/6w = 36/w = l

    then i substitute

    (36/w)^3 = maximum

    do you guys follow? in this particular case, i get some weird numbers in the end, so i think my approach didn't work in this problem?
    Try drawing the net of your box first.

    I think you'll find that if the box has a square base, then it's side lengths are l, but the height of the box is not necessarily l. We'll call it something else, h (for height).

    So TSA = 2l^2 + 4lh

    And since TSA = 216\,\textrm{ft}^2, this means

    2l^2 + 4lh = 216

    4lh = 216 - 2l^2

    h = \frac{54}{l} - \frac{l}{2}.


    Now, the volume of the box is

    V = l^2h

     = l^2\left(\frac{54}{l} - \frac{l}{2}\right)

     = 54l - \frac{l^3}{2}.


    To maximise the volume, find its derivative, set equal to 0, solve for the length, solve for the height.
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