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Thread: stationary point on a surface

  1. #1
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    stationary point on a surface

    $\displaystyle z= 3x^2 + 7x + 2y^2 - 5y -xy + 13$
    Find the stationary point and determine the nature of the point(g).
    My question is did any step im do wrongly?

    STATIONARY POINT
    $\displaystyle \frac{\partial z}{\partial x} = 6x + 7 - y$

    $\displaystyle \frac{\partial z}{\partial y} = 4y - 5 - x$
    $\displaystyle \frac{\partial z}{\partial x} = 0 ,\frac{\partial z}{\partial y} = 0$
    $\displaystyle 6x + 7 - y = 0$
    $\displaystyle 4y - 5 - x = 0$
    Then x and y value is:
    $\displaystyle x = \frac{-3}{19}$
    $\displaystyle y = \frac{23}{19}$
    Substitute x and y into original equation(z), then get $\displaystyle z=\frac{3263}{361}$

    stationary point is at ($\displaystyle \frac{-3}{19}$,$\displaystyle \frac{23}{19}$,$\displaystyle \frac{3263}{361}$)

    Now i want find the nature of the point(g):
    $\displaystyle g = (\frac{\partial^2 z}{\partial x^2})(\frac{\partial^2 z}{\partial y^2}) -(\frac{\partial^2 z}{\partial x\partial y} )^2 $
    $\displaystyle = 6(4)-(-1)^2$
    $\displaystyle =23>0$
    Stationary point is a minimum point
    Last edited by wkn0524; May 31st 2010 at 01:07 AM.
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  2. #2
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    Quote Originally Posted by wkn0524 View Post
    $\displaystyle z= 3x^2 + 7x + 2y^2 - 5y -xy + 13$
    Find the stationary point and determine the nature of the point(g).
    My question is did any step im do wrongly?

    STATIONARY POINT
    $\displaystyle \frac{\partial z}{\partial x} = 6x + 7 - y$

    $\displaystyle \frac{\partial z}{\partial y} = 4y - 5 - x$
    $\displaystyle \frac{\partial z}{\partial x} = 0 ,\frac{\partial z}{\partial y} = 0$
    $\displaystyle 6x + 7 - y = 0$
    $\displaystyle 4y - 5 - x = 0$
    Okay so far!

    Then x and y value is:
    $\displaystyle x = \frac{-3}{19}$
    $\displaystyle y = \frac{23}{19}$
    Not even close! Solve those equations again!

    Substitute x and y into original equation(z), then get $\displaystyle z=\frac{3263}{361}$

    stationary point is at ($\displaystyle \frac{-3}{19}$,$\displaystyle \frac{23}{19}$,$\displaystyle \frac{3263}{361}$)

    Now i want find the nature of the point(g):
    $\displaystyle g = (\frac{\partial^2 z}{\partial x^2})(\frac{\partial^2 z}{\partial y^2}) -(\frac{\partial^2 z}{\partial x\partial y} )^2 $
    $\displaystyle = 6(4)-(-1)^2$
    $\displaystyle =23>0$
    Stationary point is a minimum point
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  3. #3
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    Ouch, i knew my mistakes...i did a wrong common factor for x and y.

    $\displaystyle x = -1$
    $\displaystyle y = 1$

    Thanks ya
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