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swin2013 Group TitleBest ResponseYou've already chosen the best response.0
A 216m^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

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
dw:1354762359854:dw
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
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

scarydoor Group TitleBest ResponseYou've already chosen the best response.0
That looks alright to me.
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
but the second derivative i have probs with. Like finding the dimensions
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
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

scarydoor Group TitleBest ResponseYou've already chosen the best response.0
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

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
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

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
A'' = 216/y?
 one year ago

scarydoor Group TitleBest ResponseYou've already chosen the best response.0
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

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
OOOOHHHH ok.... I tried to derive it while it's a fraction it obviously didn't work hahaha
 one year ago

scarydoor Group TitleBest ResponseYou've already chosen the best response.0
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

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
Okkk, so that makes more sense that at A''(12) its positive, therefore its a min!
 one year ago

swin2013 Group TitleBest ResponseYou've already chosen the best response.0
ok so when i find the dimensions, i would have to find x?
 one year ago
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