Answer to Question #279186 in Calculus for Abc

Question

Find critical points of f (x,y) = 1/4x^2 -4xy^2 - 2x^2 +8y^2

Answer 1

$$f (x,y) = \dfrac{1}{4x^2} -4xy^2 - 2x^2 +8y^2$$

$x\not=0$

$$\dfrac{\partial f}{\partial x}=-\dfrac{1}{2x^3}-4y^2-4x$$

$$\dfrac{\partial f}{\partial y}=-8xy+16y$$

Find critical point(s)

$$\begin{cases} \dfrac{\partial f}{\partial x}=0 \\ \\ \dfrac{\partial f}{\partial x}=0 \end{cases}=>\begin{cases} -\dfrac{1}{2x^3}-4y^2-4x=0 \\ \\ -8xy+16y=0 \end{cases}$$

$$y=0=>-\dfrac{1}{2x^3}-0-4x=0$$

$$-\dfrac{1}{2x^3}=4x$$

$$8x^4=-1, No\ solution$$

$$x=2=>-\dfrac{1}{2(2)^3}-4y^2-4(2)=0$$

$$4y^2=-\dfrac{129}{16}, No\ solution$$

There are no critical points.