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

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SUMMARY

The discussion focuses on determining the nature of critical points for two functions using the Second Partial Derivative Test. For the function F=-1+4(e^x-x)-5xsin(y)+6y² at the point (0,0), the analysis concludes it is a minimum, as FxxFyy-Fxy²=23, with both Fxx and Fyy being positive. For the function F=(x²-2x)cos(y) at the stationary point (1, π), it is determined to be a maximum, with FxxFyy-Fxy²=2, despite both Fxx and Fyy being negative. The discussion confirms the correct application of the test and emphasizes the importance of checking first derivatives.

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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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