Maximum

http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-1689.jpg

圖片參考：http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-1689.jpg


2012-05-03 20:52:45 補充：
df/dx = (y+z)(z+u)(u+b)/yzu * (1 - ay/x^2) = 0=> x=sqrt(ay)
d^2f/dx^2 = (y+z)(z+u)(u+b)/yzu * (2ay/x^3) > 0 => minimum

2012-05-03 20:52:53 補充：
Alternativey (y+z)(z+u)(u+b)/yzu * (x+a)(x+y)/x
A = (x+a)(x+y)/x = x + (a+y) +ay/x = [sqrt(x) - sqrt(ay)/sqrt(x)]^2 + [sqrt(y) + sqrt(a)]^2
Holding y,z, and u constant min occurs when x = sqrt(qy)
Likewise for y,z and u to get the other 3 conditions

2012-05-03 20:53:49 補充：
typo: when x = sqrt(ay)

2012-05-04 14:45:24 補充：
I made a fundamental mistake in my original solution. I update my solution as:
http://i1090.photobucket.com/albums/i376/Nelson_Yu/int-358.jpg

3. How do you know that the maximum is not on the boundary?

2012-05-04 06:18:32 補充：
Please note a sufficient condition for local minimum is that the Hessian matrix is positive definite. It is not enough to check only the second own partial derivatives!

1. {a,x,y,z,u,b} forms a geometric sequence.
2. How do you know that the answer is a maximum rather than minimum?

2012-05-04 13:47:58 補充：
You should check that the second derivative matrix is positive or negative definite
at the indicated point.