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swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0A 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?

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0Primary (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

scarydoor
 2 years ago
Best ResponseYou've already chosen the best response.0That looks alright to me.

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0but the second derivative i have probs with. Like finding the dimensions

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0and if the second derivative turns out negative, doesn't that mean its a max? but they're asking for a min... so idk :O

scarydoor
 2 years ago
Best ResponseYou've already chosen the best response.0My 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...

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0aww 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?

scarydoor
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry. 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.

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

scarydoor
 2 years ago
Best ResponseYou've already chosen the best response.0Also 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).

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

swin2013
 2 years ago
Best ResponseYou've already chosen the best response.0ok so when i find the dimensions, i would have to find x?
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