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Math Help - Basic Application of Differentiation Question

  1. #1
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    Basic Application of Differentiation Question

    I think it's basic I'm studying it at high school and I'm led to believe that this shouldn't be that difficult!

    The question reads: A rectangular box is to be made to the following requirements:
    1) The length must be one and a half times the width. 2) The twelve edges must have a total length of 6m.
    Find the dimensions of the box that meets these requirements and that maximize the volume.

    I have the answer at the back of the book if that'd help out anyone? Thanks for your help I really appreciate it.

    My (failed) attempts so far: (Math's isnt my forte)
    1.5(4w) + 4w +4h = 6m
    then got it down to 1w = (6-4h)/10

    Subbed it into the formula I made: 1.5((6-4h)/10) + 4((6-4h)/10) +4h = 6 but it ended up saying x=x on my calc. I am lost!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by camjsiv View Post
    I think it's basic I'm studying it at high school and I'm led to believe that this shouldn't be that difficult!

    The question reads: A rectangular box is to be made to the following requirements:
    1) The length must be one and a half times the width. 2) The twelve edges must have a total length of 6m.
    Find the dimensions of the box that meets these requirements and that maximize the volume.

    I have the answer at the back of the book if that'd help out anyone? Thanks for your help I really appreciate it.

    My (failed) attempts so far: (Math's isnt my forte)
    1.5(4w) + 4w +4h = 6m
    then got it down to 1w = (6-4h)/10

    Subbed it into the formula I made: 1.5((6-4h)/10) + 4((6-4h)/10) +4h = 6 but it ended up saying x=x on my calc. I am lost!
    V=w.h.l=1.5w^2h

    The edge length constraint is:

    4w+4l+4h=4w+6w+4h=10w+4h=6

    So substituting h=(6-10w)/4 into the volume gives:

    V=1.5w^2(6-10w)/4

    To find the width corresponding to maximum volume solve:

    \frac{d}{dw}V=0

    for w and then choose the solution corresponding to maximum volume and obtain the other dimensions from that.

    CB
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