Second-Partials Test

[suicoted]

I'm stuck on this question, finding critical points. Particularly, I got stuck on how to solve the systems of equations.

1. Find all critical points of the function g(x,y) = 6x^2y - 12xy + y^3 and classify them. Show exactly how you are using the second derivative test for functions of two variables. The "y" in the first term is now raised to any power. It comes after x squared.

gx = 12xy - 12y = 0
gy = 6x^2 - 12x + 3y^2 = 0 <- now I'm kinda stuck.

The answer in the book says there are four critical points. Thanks.

 

[soroban]

Hello, suicoted!

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1. Find all critical points of the function: \(\displaystyle \,g(x,y)\:=\:6x^2y\,-\,12xy\,+\,y^3\) and classify them.
Show exactly how you are using the second derivative test for functions of two variables.

\(\displaystyle g_x\:=\:12xy\,-\,12y\:=\:0\)

\(\displaystyle g_y\:=\:6x^2\,- 12x\,+\,3y^2\:=\:0\) . . . now I'm kinda stuck.

The answer in the book says there are four critical points.

We could solve it by Substitution, but I see another way.

The first equation is: \(\displaystyle 12y(x\,-\,1)\:=\:0\)
\(\displaystyle \;\;\)which is true if: \(\displaystyle \,(a)\;\,y\,=\,0\,\) or \(\displaystyle \,(b)\;x\,=\,1\)


(a) If \(\displaystyle y\,=\,0\), the second equation becomes: \(\displaystyle \,6x^2\,-\,12x\:=\:0\)
\(\displaystyle \;\;\;\)and we have: \(\displaystyle \,6x(x\,-\,2)\:=\:0\;\;\Rightarrow\;\;x\,=\,0,\:2\)

\(\displaystyle \;\;\;\)Critical points: \(\displaystyle \,(0,\,0),\;(2,\,0)\)


(b)If \(\displaystyle x\,=\,1\), the second equation becomes: \(\displaystyle \,6\,-\,12\,+\,3y^2\:=\:0\)
\(\displaystyle \;\;\;\)and we have: \(\displaystyle \,3y^2\,=\,6\;\;\Rightarrow\;\;y^3\,=\,2\;\;\Rightarrow\;\;y\,=\,\pm\sqrt{2}\)

\(\displaystyle \;\;\;\)Crtitical points: \(\displaystyle \,(1,\,\sqrt{2}),\;(1,\,-\sqrt{2})\)


Can you finish it now?

 

[suicoted]

Yeah, thanks!
Now, I will just use d: fxx(x,y)fyy(x,y)-[fxy(x,y)]^2 for each of the critical points to see what the test gives.