# Find and classify all the critical points for f(x,y)=3x^2y-y^3-3x^2+2

Question
Analyzing functions
Find and classify all the critical points for $$\displaystyle{f{{\left({x},{y}\right)}}}={3}{x}^{{2}}{y}-{y}^{{3}}-{3}{x}^{{2}}+{2}$$

2021-03-05
Let z=f(x). Differentiating wrt x and y:
$$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{x}\right.}}={6}{x}{y}+{3}{x}^{{2}}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}-{3}{y}^{{2}}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}-{6}{x},$$
$$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{y}\right.}}={6}{x}{y}\frac{{\left.{d}{x}\right.}}{{\left.{d}{y}\right.}}+{3}{x}^{{2}}-{3}{y}^{{2}}-{6}{x}\frac{{\left.{d}{x}\right.}}{{\left.{d}{y}\right.}}.$$
$$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{x}\right.}}={3}{\left({x}^{{2}}-{y}^{{2}}\right)}\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}+{6}{x}{\left({y}-{1}\right)},$$
$$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{y}\right.}}={6}{x}{\left({y}-{1}\right)}\frac{{\left.{d}{x}\right.}}{{\left.{d}{y}\right.}}+{3}{\left({x}^{{2}}-{y}^{{2}}\right)}.$$
The general form is: $$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{t}\right.}}={6}{x}{y}\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}+{3}{x}^{{2}}\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}-{3}{y}^{{2}}\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}-{6}{x}\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}$$, where t is a parameter such that x=g(t), y=h(t) and z=j(t), where g, h and j are functions of t.
When $$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{x}\right.}}={0}$$ or $$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{y}\right.}}={0}$$ there is a critical point: $$\displaystyle\frac{{\left.{d}{y}\right.}}{{\left.{d}{x}\right.}}={2}{x}\frac{{{y}-{1}}}{{{y}^{{2}}-{x}^{{2}}}}.$$
The general form is $$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{t}\right.}}={6}{x}{\left({y}-{1}\right)}\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}+{3}{\left({x}^{{2}}-{y}^{{2}}\right)}\frac{{\left.{d}{y}\right.}}{{\left.{d}{t}\right.}}={0}.$$
So x=y=1 is a critical point, x=y=0 is another, x=-1 and y=1 is another.
When x=y=1+d, where d is very small, $$\displaystyle\frac{{\left.{d}{z}\right.}}{{\left.{d}{t}\right.}}={6}{d}{\left({1}+{d}\right)}\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}+{0}$$ which is positive when d and dx/dt are both positive or both negative. This suggests a minimum at x=y=1. If x=y=d, $$\displaystyle{6}{d}{\left({d}-{1}\right)}\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}{<}{0}$$</span>, a maximum, when d and dx/dtZSK are both positive or both negative. When $$\displaystyle{x}=-{1}+{d}{\quad\text{and}\quad}{y}={1}+{d},{6}{d}{\left({d}-{1}\right)}\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}{<}{0}$$</span>, another maximum. If d and $$\displaystyle\frac{{\left.{d}{x}\right.}}{{\left.{d}{t}\right.}}$$ are of opposite signs, the critical points are inverted.

### Relevant Questions

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