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Homework Help: Finding extreme values by partial differentiation.

  1. Aug 3, 2011 #1
    1. The problem statement, all variables and given/known data
    I sat my calculus exam at the end of june and of the questions on the paper required us to find the extreme values of the following equation:

    [itex]g\left(x,y\right) = \left(x-1\right)\left(y-1\right)\left(x^{2} + y^{2} -2\right)[/itex]

    3. The attempt at a solution
    So i get:

    [itex]\frac{\partial g}{\partial x} = 0 = \left(y-1\right)\left(3x^{2}-2x+y^{2} -2\right)[/itex]

    [itex]\frac{\partial g}{\partial y} = 0 = \left(x-1\right)\left(3y^{2}-2y+x^{2} -2\right)[/itex]

    From this I get 3 critical points [itex]\left(1,1\right), \left(1,-1\right), \left(-1,1\right)[/itex].

    However in the answer sheet there are three more, but for the life of me I can't figure out how to get them. Is there a special trick that I'm missing out on? Any help would be much appreciated!

    Final 3 points are included below for those who want to know them

    [itex]\left(-\frac{1}{2},-\frac{1}{2}\right)[/itex]
    [itex]\left(\frac{1}{2}-\frac{1}{\sqrt[]{2}},\frac{1}{2}+\frac{1}{\sqrt[]{2}}\right)[/itex]
    [itex]\left(\frac{1}{2}+\frac{1}{\sqrt[]{2}},\frac{1}{2}-\frac{1}{\sqrt[]{2}}\right)[/itex]
     
  2. jcsd
  3. Aug 3, 2011 #2

    ehild

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    The other solutions come from the system of equations

    3x2-2x+y^2-2=0 and 3y2-2y+x^2-2=0.

    ehild
     
  4. Aug 3, 2011 #3
    That's what I assumed to be the case, but unless I am doing something wrong I end up with

    [itex]\left(x-\frac{1}{2}\right)^2 - \left(y-\frac{1}{2}\right)^2 = 0[/itex]

    So the point is [itex]\left(\frac{1}{2},\frac{1}{2}\right)[/itex]

    Thanks for the quick reply though!
     
  5. Aug 3, 2011 #4

    SammyS

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    How did you solve [itex]\displaystyle \left(x-\frac{1}{2}\right)^2 - \left(y-\frac{1}{2}\right)^2 = 0\,?[/itex] There are more solutions than you gave.
    [itex]\displaystyle \left(x-\frac{1}{2}\right)^2 = \left(y-\frac{1}{2}\right)^2 [/itex]

    taking the square root of both sides gives:
    [itex]\displaystyle \left(x-\frac{1}{2}\right) = \pm\left(y-\frac{1}{2}\right) [/itex]
    Can you solve that. The solution set is the union of two lines.
     
  6. Aug 3, 2011 #5
    Wow yeah thanks, need to brush up on my more basic maths again I think. Thanks for the help from both of you! That question has been bugging me for a few weeks now =P.
     
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