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### Course: Multivariable calculus>Unit 3

Lesson 3: Optimizing multivariable functions

# Second partial derivative test example, part 2

Continuing the worked example from the previous video, now classifying each critical point. Created by Grant Sanderson.

## Want to join the conversation?

• at , when Grant compute the mixed partial derivative, he doesn't multiply sqrt(3) by 6: (Df/Dyx)=6x = 6*sqrt(3) = sqrt(108), so FxFy-Fxy^2 = -108
I understand that in this case, the magnitude of the mixed partial Derivative didn't matter since the other term =0, but is there another reason he didn't compute or might've he forgotten?
• No. Chiarandini's right. All results are -108 except for one that's 36>0. I think Grant's very stressed doing these videos! That's too bad because he's doing an awesome job otherwise.
• Wouldn't (0,0) also be the global maximum because it is the only maximum?
• No, because it is a local maximum only. Take the graph of y=x^3-x and compute the local maximum and draw the graph (with wolframalpha or so). Then you can easily see that only one local maximum exists, but that this only local maximum clearly is not the global maximum. The same principle applies to multivariable functions. If the function goes towards infinity at some point the gradient never gets to 0 because the function-value keeps getting higher and higher (towards infinity) and we only "see" it when the gradient is 0 (because that's what we calculate). I hope you got my explanation, my english isn't that well :X
• at should it be -6*6*3=-108?, since the second derivative xy=6x. why 6 was missed?
• because the x value for that critical point (0,-2) was 0, so we sub in 0 into the mixed derivative, 6x = 6(0) = 0. The video is correct.
• Hi! Thanks for sharing this new awesome MultiVariable Calculus course. Are you going to upload some video when the Second Partial Derivative Test is equal to zero? Last exam I had something like: find a,b,c Real Numbers, such that ax-by-2+2xy+cx^2-y^2. How should I proceed when I see this? Thanks!
• In this example what if fxx were positive and fyy is negative, would that be a min or max
• It would be neither, it would be a saddle point. You can't have a max or a min if fxx and fyy have different signs, and that's true for any function, not just this one.
• At , Grant forgot to write the mixed partial derivative as d^2F in the numerator, right?
• I initially thought that the 2nd-derivative test was simply the determinant of the Hessian; when the top right and bottom left terms of the Hessian are the same, then it is the same as the final term in the 2nd-derivative test. But in this case (if I've done my math correctly), the Hessian's top right and bottom left terms are not equal (6x and 6y). In this case, the determinant of the Hessian is NOT the same as the 2nd-derivative test.
Am I mistaken? Are there any connections between the determinant of the Hessian and the topography of the graph?