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X = p * y. Find x, y when both are as close to 1 as possible

  1. Mar 20, 2015 #1
    Knowing that:

    x = p * y where p = (0, 1] and x, y = (0, 2]

    find x and y in such a way that both of them be as close as possible to 1 simultaneously.

    I guess that the solution is x = 1/y = sqrt(p) but I do not know how to prove it.
     
  2. jcsd
  3. Mar 20, 2015 #2

    Mark44

    Staff: Mentor

    What does p = (0, 1] mean? What does x, y = (0, 2] mean? Does the first one mean ##x \in (0, 1]##?
    Can you pick p = 1, and then let x = y = 1?
     
  4. Mar 20, 2015 #3
    - Yes, x,y=(0,2] means x∈(0,2] and y∈(0,2].
    - p is a given constant. In my specific case p = 0.457.
     
  5. Mar 20, 2015 #4

    Mark44

    Staff: Mentor

    Since p is a constant, then for a given x value, the y value is y = x/p = x/.457 .
    If you let w = (x - 1)2 + (x/.457 - 1)2, you're looking for the value of x that minimizes w.

    Edited above to fix my earlier mistake.
     
    Last edited: Mar 21, 2015
  6. Mar 20, 2015 #5
    I believe it should be w(x) = (x - 1)^2 + (x/p - 1)^2 = minim which has the solution:
    x = p(p+1)/(p^2+1)
    It appears that x is not sqrt(p).
     
  7. Mar 21, 2015 #6

    Mark44

    Staff: Mentor

    Yes, I switched p and x due to a silly mistake on my part. I have edited my earlier post.
     
  8. Mar 21, 2015 #7
    I want now to minimize ## w(x) = (x - 1)^2 + (y - 1)^2 ## knowing that ## px^3+qx^2y+rxy^2+ty^3=0 ## but without solving the equation to obtain y as a function of x and then replace it in ## w(x) ## because this appears to be impractical. Is it possible?
     
  9. Mar 23, 2015 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Use "Lagrange multipliers". Since [itex]w(x,y)= (x- 1)^2+ (y- 1)^2[/itex], [itex]\nabla w= 2(x- 1)\vec{i}+ 2(y- 1)\vec{j}[/itex]. Writing the constraint as [itex]f(x,y)= px^3+ qx^2y+ rxy^2+ ty^3[/itex], [itex]\nabla f= (3px^2+ 2qxy+ ry^2)\vec{i}+ (qx^2+ 3rxy+ 3ty^3)\vec{j}[/itex]. At a max or min of w, with constraint f, those two gradient vectors must be parallel. That is, there must exist a number, [itex]\lambda[/itex] (the "Lagrange Multiplier") such that [itex]2(x- 1)\vec{i}+ 2(y- 1)\vec{j}[/itex][itex]= \lambda[(3px^2+ 2qxy+ ry^2)\vec{i}+ (qx^2+ 3rxy+ 3ty^3)]\vec{j}[/itex].

    That gives the equations [itex]2(x- 1)= \lambda (3px^2+ 2qxy+ ry^2)[/itex] and [itex]2(y- 1)= \lambda(qx^2+ 3rxy+ 3y^3)[/itex] which, together with the constraint, give three equations to solve or x, y, and [itex]\lambda[/itex]. Since a value for [itex]\lambda[/itex] is not part of the solution to this problem, I find that it is often simplest to first eliminate [itex]\lambda[/itex] by dividing one equation by the other. Dividing the first of those two equations by the second,
    [tex]\frac{2(x- 1)}{2(y- 1}= \frac{3px^2+ 2qxy+ ry^2}{qx^2+ 2rxy+ 3ty^3}[/tex].

    That is, of course, equivalent to [tex](x- 1)(qx^2+ 2rxy+ 3ty^3)= (y- 1)(3px^2+ 2qxy+ ry^2)[/tex].
     
  10. Mar 24, 2015 #9
    Yes, the method of Lagrange is a good idea because numeric solutions for x and y can be found with Mathcad if the parameters are known. Thank you all for your help.
     
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