SOLUTION: Topic: Maxima and Minima for functions of 2 variables: Question: Find the critical points of the question below by determining whether it is a saddle point, relative maximum or

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Question 401556: Topic: Maxima and Minima for functions of 2 variables:
Question: Find the critical points of the question below by determining whether it is a saddle point, relative maximum or minimum:
f(x,y) = xy(1-x-y)
Thank-you for your assistance.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%2Cy%29+=+xy%281-x-y%29+++=+xy+-+x%5E2y-xy%5E2
==> f%5Bx%5D+=+y-2xy+-+y%5E2+=+y%281-2x-y%29+=+0, and
f%5By%5D+=+x-2xy+-+x%5E2+=+x%281-x-2y%29+=+0.
Solving the preceding system above, we get the following c.p.'s
(i) (0,0)
{ii) (1,0)
(iii) (0,1)
(iv) (1/3, 1/3)
Finding the 2nd order partial derivatives:
f%5Bxx%5D+=+-2y
f%5Byy%5D+=+-2x, and
f%5Bxy%5D+=+1-2x+-+2y+=+f%5Byx%5D
(i) At (0,0), f%5Bxx%5D+=+0
f%5Byy%5D+=+0, and f%5Bxy%5D+=+1
D = f%5Bxx%5D%2A+f%5Byy%5D+-+%28f%5Bxy%5D%29%5E2+=+0+%2A+0+-+1+=+-1+%3C+0.
Therefore there is a saddle point at (0,0).

(ii) At (1,0), f%5Bxx%5D+=+0
f%5Byy%5D+=+-2, and f%5Bxy%5D+=+-1
D = f%5Bxx%5D%2A+f%5Byy%5D+-+%28f%5Bxy%5D%29%5E2+=+0+%2A+-2++-+1+=+-1+%3C+0.
Therefore there is a saddle point at (1,0)

(iii) At (0,1), f%5Bxx%5D+=+-2
f%5Byy%5D+=+0, and f%5Bxy%5D+=+-1
D = f%5Bxx%5D%2A+f%5Byy%5D+-+%28f%5Bxy%5D%29%5E2+=+-2+%2A+0+-+1+=+-1+%3C+0.
Therefore there is saddle point at (0,1).

(iv) At (1/3, 1/3), f%5Bxx%5D=+-2%2F3
f%5Byy%5D+=+-2%2F3, and f%5Bxy%5D+=+-1%2F3
D = .
Therefore there is a relative maximum at (1/3, 1/3).