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Thread: Stationary Point Question?

  1. #1
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    Stationary Point Question?

    Hi
    Can someone tell me if i have done the following question correctly because according to the book's answers this is incorrect.

    For the surface given $\displaystyle z=4x^3-12x+y^2+8y+8$
    Find the stationary point or points and determine the nature of any such point.

    $\displaystyle \frac{dz}{dx} = 0$
    $\displaystyle \frac{dz}{dy} = 0$

    $\displaystyle 0=12x^2-12$

    $\displaystyle x=1$

    $\displaystyle 0=2y+8$

    $\displaystyle y=-4$

    $\displaystyle z=-16$

    $\displaystyle \frac{\partial^2 z}{\partial x^2} = 12x$

    $\displaystyle \frac{\partial^2 z}{\partial y^2} = 2$

    $\displaystyle \frac{\partial^2 z}{\partial x \partial y } = 0$

    because there is still $\displaystyle 12x$ i differential again

    $\displaystyle \frac{\partial^3 z}{\partial x^3} = 12$

    $\displaystyle \frac{\partial^3 z}{\partial y^3} = 0$

    $\displaystyle \frac{\partial^3 z}{\partial x \partial y } = 0$

    $\displaystyle g= 12*0-0$

    $\displaystyle g=0$

    therefore i get saddle point according to the statement

    Points (1,-4,-16) $\displaystyle \frac{\partial^2 z}{\partial x^2}$ > 0 & g > 0

    answers is (1,-2,-8) is a local minmum; (-1,-2, 8) is a saddle point

    how do i find the minmum point?

    P.S
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  2. #2
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    Quote Originally Posted by Paymemoney View Post
    Hi
    Can someone tell me if i have done the following question correctly because according to the book's answers this is incorrect.

    For the surface given $\displaystyle z=4x^3-12x+y^2+8y+8$
    Find the stationary point or points and determine the nature of any such point.

    $\displaystyle \frac{dz}{dx} = 0$
    $\displaystyle \frac{dz}{dy} = 0$

    $\displaystyle 0=12x^2-12$

    $\displaystyle x=1$ Mr F says: You have missed the other solution to this equation ........ (I wish I had a dollar for every time this careless mistake was made).

    $\displaystyle 0=2y+8$

    $\displaystyle y=-4$ Mr F says: Correct.

    $\displaystyle z=-16$

    $\displaystyle \frac{\partial^2 z}{\partial x^2} = 12x$

    $\displaystyle \frac{\partial^2 z}{\partial y^2} = 2$

    $\displaystyle \frac{\partial^2 z}{\partial x \partial y } = 0$

    because there is still $\displaystyle 12x$ i differential again

    $\displaystyle \frac{\partial^3 z}{\partial x^3} = 12$

    $\displaystyle \frac{\partial^3 z}{\partial y^3} = 0$

    $\displaystyle \frac{\partial^3 z}{\partial x \partial y } = 0$

    $\displaystyle g= 12*0-0$

    $\displaystyle g=0$

    therefore i get saddle point according to the statement

    Points (1,-4,-16) $\displaystyle \frac{\partial^2 z}{\partial x^2}$ > 0 & g > 0

    answers is (1,-2,-8) is a local minmum; (-1,-2, 8) is a saddle point

    how do i find the minmum point?

    P.S
    The test for the nature of the critical point is given here: http://noether.uoregon.edu/~dps/242/Notes/notes21.pdf
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  3. #3
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    well if one of the points in -4 then the book's answer must be wrong.
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