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  1. #1
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    Post partial diff. min, max & saddle

    Hey All,

    how do determine the position and nature (i.e minimum, maximum or saddle point) of the stationary points of the function.

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

    this is what I have done so far
    dz/dx= 2x-4y
    dz/dy=-4x+3y^2+4

    d^2z/dxdy=-4

    how do you complete the question?

    thanks
    Last edited by dadon; May 1st 2007 at 10:14 AM.
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  2. #2
    TD!
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    The stationary points are the solutions to grad(f) = 0.
    Use your partial deratives, this gives a system in x and y.
    Use this test to determine the nature of these points.
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    sorry wrong title! thanks for the link. cool site
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  4. #4
    TD!
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    I thought the title was kind of strange

    That site is a very useful reference for math topics.
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    Hi

    so do i make the partial diff. equal to zero then find x?

    is it the same rule for normal differentation too?

    thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dadon View Post
    Hi

    so do i make the partial diff. equal to zero then find x?

    is it the same rule for normal differentation too?

    thanks
    yes, but you have to find coresponding y's as well, you need coordinates for each critical point to use in the formula
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    thanks

    what are the general rule for partial diff. to get min max and saddle

    (also in saddle the same as inflexion?)

    so you make the partial diff to zero.. to get the stationary points.

    then you differentiation again ... now this is where i get lost

    how do u find the max, min saddle from second partial diff. ?

    what is the general rule?

    thanks
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by dadon View Post
    thanks

    what are the general rule for partial diff. to get min max and saddle

    (also in saddle the same as inflexion?)

    so you make the partial diff to zero.. to get the stationary points.

    then you differentiation again ... now this is where i get lost

    how do u find the max, min saddle from second partial diff. ?

    what is the general rule?

    thanks
    the website Ticbol posted tells you everything you need to know:

    find fx and fy and set them equal to zero to find the critical points.
    find also fxx and fyy and fxy.

    FOR EACH CRITICAL POINT, evaluate fxx, fyy, fxy and plug them into the formula D = fxx*fyy - (fxy)^2, and the classifications are as follows:

    if D > 0 and fxx> 0 at the specific point, it is a relative min
    if D > 0 and fxx< 0 at the specific point, it is a relative max
    if D < 0 at the specific point, it is a saddle point
    if D = 0, you need another test, this one is inconclusive
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  9. #9
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    Hello, dadon!

    Have you never done one of these before?


    Determine the position and nature (i.e minimum, maximum or saddle point)
    of the stationary points of the function: .f(x,y) .= .x - 4xy + y + 4y
    Find the two first partial derivatives, equate to zero, and solve the system.

    . . f
    x .= .2x - 4y .= .0 - - - - . [1]
    . . f
    y .= .-4x + 3y + 4 .= .0 .[2]

    From [1], we have: .x = 2y . [3]

    Substitute into [2]: .-4(2y) + 3y + 4 .= .0

    . . We have a quadratic: .3y - 8y + 4 .= .0

    . . which factors: .(3y - 2)(y - 2) .= .0

    . . and has roots: .y .= .2/3, 2

    Substitute into [3]: .x .= .4/3, 4


    The stationary points are: .(4/3, 2/3, 32/27) and (4, 2, 0)

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Second Partials Test

    We need: .D .= .(f
    xx)(fyy) - (fxy)


    We have: .f
    xx = 2, . fyy = 6y, . fxy = -4

    Hence: .D .= .(2)(6y) - (-4) .= .12y - 16 .= .4(3y - 4)


    At (4/3, 2/3, 32/27): .D .= .4[3(2/3) - 4] .= .-8 . . . negative: saddle point

    At (4, 2, 0): .D .= .4(32 - 4) .= .+8 . . . positive: max or min
    . . f
    xx = 2 . . . positive: concave up . . . minimum

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  10. #10
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    That's great guys thanks! sorry that site is good but was abit hard for me to understand.
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