Critical Points

**Chaobunny** · March 30th 2011, 03:36 PM

So I have $f(x,y)=x^3y+12x^2-8y$ and I need to find the maxima and minima. I calculated that $\bigtriangledown f = <3yx^2+24x,x^3-8>$ , and by setting both to zero I figured out that there are critical points where $x = 0, x = -8/y, x=+-2$ . Here's where I'm running into problems. I know how to find the x values, but I'm confused about how to find the corresponding y values for these critical points. For the first one, clearly y=0 when x = 0, but I'm not sure of the others because plugging them in to the 2 equations doesn't give me anything definite. In cases like this, what is the best way to find the corresponding y values of the critical points?

**Soroban** · March 30th 2011, 05:36 PM

Hello, Chaobunny!

$f(x,y)\:=\:x^3y+12x^2-8y.\;\;\text{ Find the maxima and minima.}$

$\text{I found that }\nabla f \:=\:\langle 3yx^2+24x,\,x^3-8\rangle$

$\text{Setting both to zero, I found that there are critical points}$
. . $\text{where: }\:x = 0,\:x = -\frac{8}{y},\:x= \pm 2$ . . How?

We have: . $\begin{Bmatrix}f_x &=& 3x^2y + 24x &=&0 & [1] \\ f_y &=& x^3-8 &=& 0 & [2] \end{Bmatrix}$

From [2], we have: . $x^3 \:=\:8 \quad\Rightarrow\quad x \;=\:2$

Substitute into [1]: . $12y + 48 \:=\:0 \quad\Rightarrow\quad y \:=\:\text{-}4$

We have one critical point: . $(2,\,\text}-}4,\,48)$

We have: . $\begin{Bmatrix}f_{xx} &=& 6xy + 24 \\ f_{yy} &=& 0 \\ f_{xy} &=& 3x^2 \end{Bmatrix}$

$D \;=\;(f_{xx})(f_{yy}) - (f_{xy})^2 \;=\;(6xy+24)(0) - (3x^2)^2 \;=\;-9x^4$

At $(2, \text{-}4, 48)$ , we have: . $D \:=\:-9(2^4) \:=\:-144$

Since $\,D$ is negative, there is a saddle point at $(2, \text{-}4, 48).$

**Chaobunny** · March 30th 2011, 05:50 PM

Thank you very much! So, as a clarification, if there isn't an obvious corresponding y value, then is there no critical point at that location?

**HallsofIvy** · March 31st 2011, 05:03 AM

I have no idea what you mean by "an obvious corresponding y value". The point is that both equations, $3yx^2+ 24x= 0$ and $x^3- 8= 0$ must be true at a critical point. The only solution to $x^3- 8= 0$ is x= 2 (NOT x= -2. If x= -2, $x^3- 8= -8- 8= -16$ , not 0) and, putting that back into the first equation, $12y+ 48= 0$ so that $y= -4$ .

The fact that x= 0 satisfies the first equation is irrelevant- it does not satisfy the second equation for any value of y.

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