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swin2013

  • 2 years ago

Optimization question?

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  1. swin2013
    • 2 years ago
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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?

  2. swin2013
    • 2 years ago
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    |dw:1354762359854:dw|

  3. swin2013
    • 2 years ago
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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

  4. swin2013
    • 2 years ago
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    @zepdrix

  5. scarydoor
    • 2 years ago
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    That looks alright to me.

  6. swin2013
    • 2 years ago
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    but the second derivative i have probs with. Like finding the dimensions

  7. swin2013
    • 2 years ago
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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

  8. scarydoor
    • 2 years ago
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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...

  9. swin2013
    • 2 years ago
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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?

  10. swin2013
    • 2 years ago
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    A'' = -216/y?

  11. scarydoor
    • 2 years ago
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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.

  12. swin2013
    • 2 years ago
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    OOOOHHHH ok.... I tried to derive it while it's a fraction it obviously didn't work hahaha

  13. scarydoor
    • 2 years ago
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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).

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

  15. swin2013
    • 2 years ago
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    ok so when i find the dimensions, i would have to find x?

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