Answer to Question #300305 in Calculus for sha

Question

Locate and classify all critical points for f(x, y) = 4y^2x + 3xy

Answer 1

Find the critical point(s)

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

$$\dfrac{\partial f}{\partial y}=8xy+3x$$

Then

$$\begin{cases} \dfrac{\partial f}{\partial x}=0 \\ \\ \dfrac{\partial f}{\partial y}=0 \end{cases}$$

$$\begin{cases} 4y^2+3y=0 \\ 8xy+3x=0 \end{cases}$$

Critical points $(0, 0), (0, -3/4)$

Point $(0,0)$

$$\dfrac{\partial^2 f}{\partial x^2}(0,0)=0$$

$$\dfrac{\partial^2 f}{\partial y^2}(0,0)=8(0)=0$$

$$\dfrac{\partial^2 f}{\partial x\partial y}(0,0)=\dfrac{\partial^2 f}{\partial y\partial x}(0,0)=8(0)+3=3$$

$$D(0,0)=\begin{vmatrix} 0 & 3 \\ 3 & 0 \end{vmatrix}=-9<0$$

Then the $f(0,0)$ is not a local maximum or minimum.

The critical point $(0,0)$ is a saddle point.

Point $(0,-3/4)$

$$\dfrac{\partial^2 f}{\partial x^2}(0,-3/4)=0$$

$$\dfrac{\partial^2 f}{\partial y^2}(0,-3/4)=8(0)=0$$

$$\dfrac{\partial^2 f}{\partial x\partial y}(0,-3/4)=\dfrac{\partial^2 f}{\partial y\partial x}(0,-3/4)$$

$$=8(-3/4)+3=-3$$

$$D(0,0)=\begin{vmatrix} 0 & -3 \\ -3 & 0 \end{vmatrix}=-9<0$$

Then the $f(0,-3/4)$ is not a local maximum or minimum.

The critical point $(0,-3/4)$ is a saddle point.