Results 1 to 4 of 4

Math Help - Partial Differentiation stationary points

  1. #1
    Junior Member
    Joined
    Apr 2010
    Posts
    59

    Partial Differentiation stationary points

    hey im pretty stuck with this, I've got quite far though, I think..

    f(x,y) = 1/3x^3 + 1/3y^3 - 4xy^2 +15y

    then df/dx =  x^2 - 4y^2
    df/dy =  y^2 - 8xy + 15

    Then to find the stationary points, we must equate the differentials to 0, and I get to the 4 points:

    (2y, -15), (2y, 8x+15), (-2y, -15), (-2y,8x+15)

    Then by using the Hessian matrix

    H = fxx fxy
    fyx fyy

    Where fxx = 2x, fyy = 2y - 8x, fyx = -8, fxy = 8y

    Then the determinant is 4xy - 16x^2 + 64y

    Then you put in the x and y co-ordinates and determine whether is a maxima or minima as < or > 0.

    But I don't have any numbers...so if anybody can help or see where I've gone wrong

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81
    Quote Originally Posted by ramdrop View Post
    hey im pretty stuck with this, I've got quite far though, I think..

    f(x,y) = 1/3x^3 + 1/3y^3 - 4xy^2 +15y

    then df/dx =  x^2 - 4y^2
    df/dy =  y^2 - 8xy + 15

    Then to find the stationary points, we must equate the differentials to 0, and I get to the 4 points:

    (2y, -15), (2y, 8x+15), (-2y, -15), (-2y,8x+15)

    Then by using the Hessian matrix

    H = fxx fxy
    fyx fyy

    Where fxx = 2x, fyy = 2y - 8x, fyx = -8, fxy = 8y

    Then the determinant is 4xy - 16x^2 + 64y

    Then you put in the x and y co-ordinates and determine whether is a maxima or minima as < or > 0.

    But I don't have any numbers...so if anybody can help or see where I've gone wrong

    Thanks
    It seems you did not do the arithmetics.
    As you have

    \displaystyle \[\left\{ \begin{array}{l}<br />
\frac{{\partial f}}{{\partial x}} = {x^2} - 4{y^2} = 0\\<br />
\\<br />
\frac{{\partial f}}{{\partial y}} = {y^2} - 8xy + 15 = 0<br />
\end{array} \right.\]

    From the second one

    \displaystyle \[{y^2} - 8xy + 15 = 0 \Rightarrow x = \frac{{{y^2} + 15}}{{8y}}\]

    We go back to the first one

    \displaystyle \[{x^2} - 4{y^2} = 0 \Rightarrow {\left( {\frac{{{y^2} + 15}}{{8y}}} \right)^2} - 4{y^2} = 0 \Rightarrow \frac{{{y^4} + 30{y^2} + 225 - 256{y^4}}}{{64{y^2}}} = 0\]

    Which gives us

    \displaystyle \[255{y^4} - 30{y^2} - 225 = 0 \Rightarrow 17{y^4} - 2{y^2} - 15 = 0 \Rightarrow (y - 1)(y + 1)(17{y^2} + 15) = 0\]

    And exact values are at our hands

    \displaystyle \[{x_1} = 2\] and \displaystyle \[{y_1} = 1\]

    \displaystyle \[{x_2} =  - 2\] and \displaystyle \[{y_2} =  - 1\]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2010
    Posts
    59
    Hm I tried subbing it in, but i re-arranged the first and went into the second, many thanks bud
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,258
    Thanks
    1798
    Quote Originally Posted by ramdrop View Post
    Hm I tried subbing it in, but i re-arranged the first and went into the second, many thanks bud
    Perhaps you simply do not understand what is meant by "solve the equations". To solve
    x^2- 4y^2= 0 and y^2- 8xy+ 15= 0
    does NOT mean to solve one equation for x as a function of y (which is what you apparently did to get something like "(2y, -15)".

    You can, as Pranas said, solve x^2- 4y^2 for x as a function of y: x= \pm 2y and then put, first, x= 2y into the second equation: y^2- 8xy+ 15= y^2- 8(2y)y+ 15= -15y^2+ 15 which reduces to 15y^2= 15 or y= \pm 1. Then, since x= 2y, x= 2(1)= 2 or x= 2(-1)= -2. Two solutions are (2, 1) and (-2, -1). Putting x= -2y into that equation instead, [tex]y^2- 8(-2y)y+ 15= 17y^2+ 15= 0 which has no real solutions.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 24th 2011, 12:35 PM
  2. finding stationary points using differentiation
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 20th 2009, 06:44 AM
  3. Stationary Points
    Posted in the Calculus Forum
    Replies: 4
    Last Post: December 3rd 2008, 01:18 PM
  4. Stationary Points, Local points of Extrema
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 19th 2008, 02:44 PM
  5. Replies: 3
    Last Post: May 5th 2006, 10:22 AM

Search Tags


/mathhelpforum @mathhelpforum