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Math Help - 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 z=4x^3-12x+y^2+8y+8
    Find the stationary point or points and determine the nature of any such point.

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

    0=12x^2-12

    x=1

    0=2y+8

    y=-4

    z=-16

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

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

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

    because there is still 12x i differential again

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

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

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

    g= 12*0-0

    g=0

    therefore i get saddle point according to the statement

    Points (1,-4,-16) \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 z=4x^3-12x+y^2+8y+8
    Find the stationary point or points and determine the nature of any such point.

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

    0=12x^2-12

    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).

    0=2y+8

    y=-4 Mr F says: Correct.

    z=-16

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

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

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

    because there is still 12x i differential again

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

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

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

    g= 12*0-0

    g=0

    therefore i get saddle point according to the statement

    Points (1,-4,-16) \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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