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Math Help - Critical Points

  1. #1
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    Critical Points

    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?
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  2. #2
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    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).

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  3. #3
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    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?
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  4. #4
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    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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