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If $f(x,y) = x^{2}-4xy+y^{3}+4y$, find and classify all local extrema.

The answer is supposed to be:

$(4,2)$ is a minimum and $(\frac{4}{3},\frac{2}{3})$ is a saddle point.

This is my work:

$$f_x=2x-4y$$ $$f_{xx}=2$$ $$f_{xy}=-4$$

$$f_y=-4x+3y^{2}+4$$ $$f_{yy}=6y$$ $$f_{yx}=-4$$$

Then, after setting $f_{x}=0$ and $f_y=0$, the points I get are $(\frac{8}{3},\frac{4}{3})$ and $(-4,-4)$.

What am I doing wrong?

juliodesa
  • 155

2 Answers2

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Partials derivatives are okay. Then you should have done $$f_x(x,y)=0\implies 2x-4y=0\implies x=2y$$ and $$f_y(x,y)=0\implies-4x+3y^2+4=0\implies3y^2-8y+4=0.$$ Here see that if you note $P=3X^2-8X+4$ then $\Delta_P=16$ then your roots are $\frac{8\pm 4}{6}$ so $y=\frac{2}{3}$ and $y=2.$ Finally $(\frac{4}{3},\frac{2}{3})$ and $(4,2)$ are your only extrema possible.

Balloon
  • 8,374
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$f(x,y)=x^2-4xy+y^3+4y$, so $\nabla f(x,y)=(2x-4y, -4x+3y^2+4)$. We look for points where $\nabla f(x,y) = (0,0)$, yielding that $x=2y$, which in the second equation yields: $$-4x+3y^2+4= 3y^2-8y+4 = 3(y-4/3)^2-16/3+4= 0 \iff (y-4/3)^2=4/9 $$

This yields that $y=\frac{4 \pm 2}{3}$, and consequently $x=\frac{8\pm 4}{3}$.

So we get the local extrema $(\frac{4}{3},\frac{2}{3})$ and $(4,2)$. To classify the local extrema, use the Hessian, and determine whether it's definite or indefinite in the given points.

Scounged
  • 937