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Math Help - Smallest area and Smallest Length of String around Box.

  1. #1
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    Smallest area and Smallest Length of String around Box.

    I have a problem!

    I have attached it to the thread, just open the attachment please.
    Attached Files Attached Files
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  2. #2
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    For starters, I think you'll find that the length of the string is actually 6x + 2y...
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  3. #3
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    thankyou for that.
    are you able to help any further?
    Is my method ok besides my mistake?
    How can i use it for the question i am stuck on, please?
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  4. #4
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    Please can anyone give me a few pointers??
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  5. #5
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    I think that in the problem you solved, the rope length is 4x + 2y, not 4x + 4y. Correspondingly, L = 4x + 2V / x^2, not 4x + V / x^2. Otherwise, I agree.

    In the first problem, the variables m, w, and h seem independent, so the rope length has to be expressed through all three. Since V = 2mwh, you can express one of m, w, h through V and the other two and so express L through V and two of m, w, h. In particular, I don't see how you can find L in terms of V and w only. Also, minimizing the rope length requires some other constraint (e.g., keeping the volume constant); otherwise, you can get L = 0 when m = w= h = 0 .
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  6. #6
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    ahh sorry i have forgotten to say that volume = 3m^3 and w = 3m
    so would you do something like this
    L = 4w + 4m + 6H
    For V = 2mwH make h the subject H = V/2mwH ??
    Sub into equation
    L = 4w + 4m + 3V/2mw
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  7. #7
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    First, I get

    L = 2(2h + 4m + 2w + 2h) = 8h + 8m + 4w. (*)

    Here 2h corresponds to the front side, 4m + 2w to the top side, and 2h to the left side.

    Second, V = 2mwh; substituting w = 3m, V = 6m^2h. Since also V = 3m^3, we get 2h = m. So, the side lengths are h, m = 2h and w = 3m = 6h. Also, V = 2mwh = 2 * 2h * 6h * h = 24h^3. So, everything can be expressed through h.

    If you still need to find L in terms of w, then from (*) I get L = 8h + 8m + 4w = (8/6)w + (8/3)w + 4w.
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  8. #8
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    ahh oh i see gosh i was way off track... Just to clarify i don't fully understand how you did that last part when finding L in terms of w and volume?

    Also, the question asks to find L in terms of w and Volume. I need V don't i?
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  9. #9
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    Since w = 6h and w = 3m, we get h = w/6 and m = w/3. So, L = 8h + 8m + 4w = 8(w/6) + 8(w/3) + 4w = (4/3 +8/3 + 4)w = 8w.

    I suggest you verify all calculations by doing them yourself without looking here because I make quite a few errors.
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  10. #10
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    Sorry to bother you further....
    THe question asks to find the Length in terms of Width and Volume

    i don't think this applies to you above solution or am i mistaken?

    (oh and yes i do verify my all the calculations i have been nutting this question out on paper myself guided by your ideas in your threads)
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  11. #11
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    THe question asks to find the Length in terms of Width and Volume
    Well, L = 8w satisfies this requirement; it just does not use volume.

    You gave two additional equations: volume = 3m^3 and w = 3m, which allowed reducing the number of independent variables. Initially, there are 5 variables: m, w, h, V, L and 2 equations: V = 2mwh and L = 8h + 8m + 4w. The number of independent variables is the total number of variables minus the number of equations (that can't be derived from each other), so initially there are 3 independent variables. When you added V = 3m^3 and w = 3m, the number of independent variables was reduced to 1.
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  12. #12
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    ok so if i was to draw a graph of v vs w how would that work?
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  13. #13
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    V = 2mwh = 2 * (w/3) * w * (w/6) = w^3 / 9. Here is the graph in WolframAlpha.
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  14. #14
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    Ah i see... Thankyou so much for your time. This will help me to practice other questions similar
    Thanks again
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