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swin2013 Group Title

Optimization question?

  • one year ago
  • one year ago

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  1. swin2013 Group Title
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    A 216-m^2 rectangular pea patch is enclosed by a fence and divided into two equal parts by fence parallel to one of the sides. What dimensions for the rectangle will require the smallest total length of fence much fence will be needed?

    • one year ago
  2. swin2013 Group Title
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    |dw:1354762359854:dw|

    • one year ago
  3. swin2013 Group Title
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    Primary (min) = P = 2x +3y Secondary = xy=216 x = 216/ y => 2(216/y) +3y = P 432/y +3y = P P' = 3 - 432/y^2 y = 12? And i'm pretty sure I did the double derivative wrong

    • one year ago
  4. swin2013 Group Title
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    @zepdrix

    • one year ago
  5. scarydoor Group Title
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    That looks alright to me.

    • one year ago
  6. swin2013 Group Title
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    but the second derivative i have probs with. Like finding the dimensions

    • one year ago
  7. swin2013 Group Title
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    and if the second derivative turns out negative, doesn't that mean its a max? but they're asking for a min... so idk :O

    • one year ago
  8. scarydoor Group Title
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    My brain isn't working today... But I've plotted the function P(y) and it has a minimum at y=12. That's not a very mathematical argument, but... I'm not thinking well today...

    • one year ago
  9. swin2013 Group Title
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    aww that's ok... that makes the both of us! hahaha well 12 is the correct y coordinate. so yes, i'm finding the minimum. But I'm not sure if my 2nd derivative is correct?

    • one year ago
  10. swin2013 Group Title
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    A'' = -216/y?

    • one year ago
  11. scarydoor Group Title
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    Sorry. You're right. http://en.wikipedia.org/wiki/Second_derivative#Second_derivative_test But P'(y) = -432 * y^(-2). So P''(y) = -2 * -432 * y^(-3) = 864 / y^3. And that is positive for y=12. So it's a local minimum.

    • one year ago
  12. swin2013 Group Title
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    OOOOHHHH ok.... I tried to derive it while it's a fraction it obviously didn't work hahaha

    • one year ago
  13. scarydoor Group Title
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    Also you found that P has a critical point only at y=12. So the global minimum can only either be at that critical point, or at the end point of your domain for y (eg, if you require y is in the interval [a, b], then you also check the value of P(a) and P(b)). So then, here, the min must be P(12).

    • one year ago
  14. swin2013 Group Title
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    Okkk, so that makes more sense that at A''(12) its positive, therefore its a min!

    • one year ago
  15. swin2013 Group Title
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    ok so when i find the dimensions, i would have to find x?

    • one year ago
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