Here's the question you clicked on:
swin2013
Optimization question?
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?
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
That looks alright to me.
but the second derivative i have probs with. Like finding the dimensions
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
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...
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?
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.
OOOOHHHH ok.... I tried to derive it while it's a fraction it obviously didn't work hahaha
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).
Okkk, so that makes more sense that at A''(12) its positive, therefore its a min!
ok so when i find the dimensions, i would have to find x?