## swin2013 2 years ago Optimization question?

1. swin2013

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

|dw:1354762359854:dw|

3. swin2013

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

@zepdrix

5. scarydoor

That looks alright to me.

6. swin2013

but the second derivative i have probs with. Like finding the dimensions

7. swin2013

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

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

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

A'' = -216/y?

11. scarydoor

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

OOOOHHHH ok.... I tried to derive it while it's a fraction it obviously didn't work hahaha

13. scarydoor

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

Okkk, so that makes more sense that at A''(12) its positive, therefore its a min!

15. swin2013

ok so when i find the dimensions, i would have to find x?