I want to confirm if I am right,

The critical points that i got from this equation

C = x4 - 8xy + 2y2 - 3

is (0,0)

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- Mar 21st 2009, 10:16 PMHakhengkimFInding Critical Point Using Partial Derivative
I want to confirm if I am right,

The critical points that i got from this equation

C = x4 - 8xy + 2y2 - 3

is (0,0)

- Mar 21st 2009, 10:47 PMSpec
(0,0) is not the only critical point.

- Mar 21st 2009, 10:49 PMHakhengkim
- Mar 21st 2009, 10:50 PMChris L T521
The function is

Thus, and

Now to find the critical points, solve and

This would mean that and .

Solving the second equation, we have

Substituting this into the first equation, we have

Thus, or .

As a result, when:

Thus, your critical points should be , and

Does this make sense? - Mar 21st 2009, 11:00 PMHakhengkim