Results 1 to 5 of 5

Math Help - FInding Critical Point Using Partial Derivative

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    16

    FInding Critical Point Using Partial Derivative

    I want to confirm if I am right,

    The critical points that i got from this equation

    C = x4 - 8xy + 2y2 - 3

    is (0,0)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    (0,0) is not the only critical point.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    16
    Quote Originally Posted by Spec View Post
    (0,0) is not the only critical point.
    Really what are the other ones . By any chance can you show me?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Hakhengkim View Post
    I want to confirm if I am right,

    The critical points that i got from this equation

    C = x4 - 8xy + 2y2 - 3

    is (0,0)
    The function is f\!\left(x,y\right)=x^4-8xy+2y^2-3

    Thus, \frac{\partial f}{\partial x}=4x^3-8y and \frac{\partial f}{\partial y}=4y-8x

    Now to find the critical points, solve \frac{\partial f}{\partial x}=0 and \frac{\partial f}{\partial y}=0

    This would mean that 4x^3-8y=0 and 4y-8x=0.

    Solving the second equation, we have 4y=8x

    Substituting this into the first equation, we have 4x^3-2(8x)=0\implies 4x^3-16x=0\implies 4x(x^2-4)=0

    Thus, x=0,~x=2 or x=-2.

    As a result, when:

    x=0\,:\, 4y=8(0)\implies y=0

    x=2\,:\, 4y=8(2)\implies y=4

    x=-2\,:\, 4y=8(-2)\implies y=-4

    Thus, your critical points should be \left(0,0\right), \left(2,4\right) and \left(-2,-4\right)

    Does this make sense?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2009
    Posts
    16
    Quote Originally Posted by Chris L T521 View Post
    The function is f\!\left(x,y\right)=x^4-8xy+2y^2-3

    Thus, \frac{\partial f}{\partial x}=4x^3-8y and \frac{\partial f}{\partial y}=4y-8x

    Now to find the critical points, solve \frac{\partial f}{\partial x}=0 and \frac{\partial f}{\partial y}=0

    This would mean that 4x^3-8y=0 and 4y-8x=0.

    Solving the second equation, we have 4y=8x

    Substituting this into the first equation, we have 4x^3-2(8x)=0\implies 4x^3-16x=0\implies 4x(x^2-4)=0

    Thus, x=0,~x=2 or x=-2.

    As a result, when:

    x=0\,:\, 4y=8(0)\implies y=0

    x=2\,:\, 4y=8(2)\implies y=4

    x=-2\,:\, 4y=8(-2)\implies y=-4

    Thus, your critical points should be \left(0,0\right), \left(2,4\right) and \left(-2,-4\right)

    Does this make sense?

    Ahhh ic thank you totally forgot about the +4y on the second partial derivative.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. trig derivative critical point
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 13th 2011, 07:32 PM
  2. Finding the critical point
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 18th 2010, 03:35 PM
  3. Replies: 0
    Last Post: November 3rd 2009, 11:18 AM
  4. Finding the critical point
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 3rd 2009, 08:27 AM
  5. Finding a Critical Point
    Posted in the Calculus Forum
    Replies: 10
    Last Post: October 22nd 2006, 10:21 AM

Search Tags


/mathhelpforum @mathhelpforum