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Homework Help: Max/Min in Calc III- Confused about relation between equations logic?

  1. Mar 20, 2012 #1
    So, I technically got this problem correct (our professor gave us the answers but not the total solution). My question is more of a logic one- an answer that I'll need to apply this concept to problems on the exam. I included it at the bottom of the post, after I explained the problem

    Find local max and min of f(x,y)=xy2-yx2+xy

    2. Relevant equations

    fx=y2-2yx+y fy=2xy-x2+x
    fxx=-2y fxy=2y-2x+1 fyy=2x

    3. The attempt at a solution

    y=0 OR y-2x+1=0

    x=0 OR 2y-x+1=0

    Plug y=0 into 2y-x+1=0

    Plug x=0 into y-2x+1=0

    Therefore, -y=x. Plug this into y-2x+1=0

    y=-1/3 x=1/3

    Critical Points:

    **(I know how to solve for min and max from here, I don't need to type it out)

    My question:
    I don't understand the relationship between the fy and fx derivatives. For instance, I understand why x=0 OR 2y-x+1=0 (from fx), but I don't understand why you would assume that plugging y=0 into 2y-x+1=0 is correct? Why does 2y-x+1=0 when y=0? Rather, why does plugging y=0 into 2y-x+1=0 give you a point on the graph, when 2y-x+1 isn't the derivative itself, only only part of the factored equation? Why don't you take into account the y factored out of (2y-x+1), since you can't factor out a zero (because you're assuming y=0). If x=0 and y=0 are both solutions, why isn't (0,0) a critical point?

    Furthermore, I understand the algebra behind plugging -y=x into y-2x+1=0 to get -1/3 and 1/3, but I would never think to do that. How do you know there is a third answer, when you already found 2 answers from the factored y-2x+1=0 (or 2y-x+1=0).

    Its just hard for me to visualize whats going on!
  2. jcsd
  3. Mar 20, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    The conditions f_x = 0 and f_y = 0 give you two equations:
    y^2-2yx+y = 0 (1)
    2xy-x^2+x = 0 (2)

    Eq (1) implies y(y - 2x + 1) = 0, so we must have either
    Case (a) y = 0; or Case (b) y - 2x + 1 = 0 --> y = 2x - 1.

    Note: at this point we have used eq (1) fully, so now look at eq (2).

    In Case (a), eq (2) gives x = x^2, so either x = 0 or x = 1; that is, we have two solutions in case (a): (x,y) = (0,0) and (x,y) = (1,0).

    In Case (b) we substitute y = 2x - 1 into eq (2), to get 2x(2x-1)-x^2+x = 0 = x(4x-2-x+1), hence x = 0 or 3x=1. When x = 0 we have y = 2*0-1 = -1 and when x = 1/3 we have y = 2/3 - 1 = -1/3. Therefore, case (b) leads to solutions (x,y) = (0,-1) and (x,y) = (1/3,-1/3).

    By breaking the analysis up into cases you can avoid getting mixed up or missing some possibilities.

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