Min, Max, or Saddle Point at (0,0) for F=?

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For part (a), the function F at the point (0,0) is determined to be a minimum based on the second partial derivative test, with Fxx, Fyy both positive and the determinant being greater than zero. In part (b), the function F at the stationary point (1, π) is identified as a maximum, as the determinant is positive but both Fxx and Fyy are negative. The discussion emphasizes the importance of ensuring that the first derivatives are zero at the stationary points. The professor's explanation was unclear, but the application of the second partial derivative test was confirmed to be correct. Overall, the analysis of both functions aligns with the established criteria for classifying critical points.
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Homework Statement



Decide between a minimum, maximum, or saddle point for:

(a) F=-1+4(ex-x)-5xsin(y)+6y2 at the point x=y=0

(b) F=(x2-2x)cos(y) with stationary point at x=1, y=pi

The professor was a bit confusing, so I did it the way I remember from Calc, but am unsure of whether I went the right direction or not... Ideas?

Homework Equations



FxxFyy-Fxy2

The Attempt at a Solution



For part a,

I got the point is at a minimum because
FxxFyy-Fxy2=23, which is greater than 0, and so are both Fxx and Fyy (4 and 12 respectively).

For part b,

I got the point is at a maximum because
FxxFyy-Fxy2=2, which is greater than 0, but both Fxx and Fyy are negative (-2 and -1 respectively).

Am I thinking correctly? Thanks!

 
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