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Math Help - Critical values of Z=(x-2)^4+(y-3)^4.

  1. #1
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    Exclamation Critical values of Z=(x-2)^4+(y-3)^4.

    Z=(x-2)^4+(y-3)^4
    plz help me 2 find out the critical and extreme values
    Last edited by mr fantastic; May 27th 2010 at 08:17 PM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by powerxxx View Post
    Z=(x-2)^4+(y-3)^4
    plz help me 2 find out the critical and extreme values
    Solve simultaneously:

    \frac{\partial z}{\partial x} = 0

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

    If you need more help, please show all your work and say exactly where you get stuck.
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  3. #3
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    Iím new... could you plz slow me the step by step process for this problem
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    Quote Originally Posted by powerxxx View Post
    I’m new... could you plz slow me the step by step process for this problem
    If you do not know how to at least get the partial derivatives, then I suggest you forget about this question for a while and go back and review what you have already been taught about partial differentiation (the question assumes that you know it).

    Otherwise, please show all your working so far and say where you are stuck.
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  5. #5
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    Quote Originally Posted by powerxxx View Post
    Iím new... could you plz slow me the step by step process for this problem
    We understand that you are new but you must show us an attempt at the problem to prove you are trying.

    Find

    f_x
    f_{xx}
    f_y
    f_{yy}
    f_{xy}

    Then use

    (f_{xx})(f_{yy}) -(f{xy})^2
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  6. #6
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    f_X=4(X-2)^3
    f_XX=12(X-2)^2
    f_XY=0

    f_Y=4(Y-3)^3
    f_YY=12(Y-3)^2
    F_YX=0

    but how i will get the extreme value
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by powerxxx View Post
    f_X=4(X-2)^3
    f_XX=12(X-2)^2
    f_XY=0

    f_Y=4(Y-3)^3
    f_YY=12(Y-3)^2
    F_YX=0

    but how i will get the extreme value
    Ok now set your 1st partial derivatives to 0

    4(x-2)^3 = 0

    x = 2

    4(y-3)^3 = 0

    y = 3

    so your critical value is (2,3)

    Now use

    (f_{xx})(f_{yy}) -(f{xy})^2 = D

    if D = 0 inconclusive
    if D > 0 max or min
    if D < 0 saddle point
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