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Maximizing a multivariate function under a constraint

  1. Jun 17, 2012 #1
    1. The problem statement, all variables and given/known data


    Consider [itex] f(x,y) = \frac{1}{x} - \frac{1}{y} [/itex]

    You need to maximize [itex] f(x,y) [/itex] given the constraint:
    [itex] x + y = 11 [/itex]


    2. Relevant equations
    I have never solved a problem like this before. In fact I made up a problem a few minutes ago.


    3. The attempt at a solution
    I can only imagine graphing f(x,y) on the x,y plane and seeing the maximum point of intersection with the plane of constraint. Actually I am not even sure if that is right.

    BiP
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 17, 2012 #2

    SammyS

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    Solve for y: [itex] x + y = 11 [/itex] .

    Plug that into [itex]\displaystyle f(x,y) = \frac{1}{x} - \frac{1}{y} [/itex], so that you have a function of x only.
     
  4. Jun 17, 2012 #3
    I see. The single-variate function of x that is obtained once I do that: what is its visual relationship to the graphs of f(x,y) and x+y=11 ?

    Is it their intersection? It would be helpful to have a visual reference for each step so I can clearly understand what's going on.

    BiP
     
  5. Jun 17, 2012 #4

    Ray Vickson

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    Say you have a problem of the form max f(x,y), subject to g(x,y) = 0, with both f and g at least once continuously differentiable. In your case
    [tex] f(x,y) = \frac{1}{x} - \frac{1}{y} \text{ and } g(x,y) = x+y-11[/tex]
    but the basic idea is more general. The constraint g=0 gives a curve C in the (x,y)-plane from which we want to select the point (x,y) giving the largest f. What conditions must hold at the optimal solution p* = (x*,y*) (assuming there IS a finite solution)? Basically, the gradient vectors [itex]Df =\nabla f [/itex] and [itex] Dg = \nabla g[/itex] must be parallel or antiparallel at p*. The reason is this: if [itex]Df[/itex] and [itex]Dg[/itex] point along different lines, there is a nonzero component of [itex]Df [/itex] that is perpendicular to [itex]Dg[/itex], so is tangent to C at the point p*; and that means that by going away from p* (but staying in C), we can make f go up a bit. That would mean that f(p*) is not the maximum. Therefore, to prevent this from happening we need the two gradient vectors to point along the same line (but maybe in opposite directions).

    If you imagine a plot of "level curves" of f and g (like the constant-altitude contours in a hiker's area map), then at the point p* the level curves of f and g are parallel---that is, they have the same tangent line.

    The geometric property above can be expressed by saying that at p* there is a scalar λ such that [itex] \nabla f = \lambda \nabla g[/itex]. The scalar is usually called the Lagrange Multiplier, and the optimization rule amounts to saying that the so-called Lagrangian
    [itex] L = f - \lambda g[/itex] must be stationary at p*; that is, we need [itex] \nabla L = 0[/itex] at p* = (x*,y*). Note that λ is not a function of x or y.

    So, let's do this for your problem:
    [tex] L = \frac{1}{x}-\frac{1}{y} - \lambda(x+y-11).[/tex]
    We have
    [tex] \frac{\partial L}{\partial x} = 0 \longrightarrow -\frac{1}{x^2} - \lambda = 0,\\
    \frac{\partial L}{\partial y} = 0 \longrightarrow +\frac{1}{y^2} - \lambda = 0.
    [/tex]
    From this, we see that [itex] x^2 = -y^2[/itex] at [itex] x = x^*,\, y = y^*,[/itex] and that can only happen if [itex] x^* = y^* = 0.[/itex] But f(0,0) is not defined, so there is no optimal solution. You can also see that this is the case by substituting y = 11-x into f and looking at a plot of the result.

    A nicer example would be to take [tex] f = \frac{1}{x} + \frac{2}{y}[/tex] and keeping the same g as before. Now you would get [tex] L = \frac{1}{x} + \frac{2}{y} - \lambda (x+y-11), [/tex] and so
    [tex] \frac{\partial L}{\partial x} = -\frac{1}{x^2} - \lambda = 0,\\
    \frac{\partial L}{\partial y} = -\frac{2}{y^2} - \lambda = 0,\\
    x+y=11,[/tex]
    so either (1)
    [tex] x = 11\sqrt{2}-11, \; y = 22 - 11\sqrt{2}, \; \lambda = -\frac{3+2\sqrt{2}}{121},[/tex]
    or (2)
    [tex] x = -11 - 11 \sqrt{2}, \; y = 22 + 11\sqrt{2}, \; \lambda = -\frac{3 - 2\sqrt{2}}{121}.[/tex]

    It turns out that point (1) is a maximum while point (2) is a minimum. In general we would need to perform second-order tests to check whether a point is a max or a min (or neither), because just looking at gradients equal to zero does not tell us whether the point is a max or a min. However, in this case we can just evaluate f at the two points to see which is which.

    Note: by max and min above, we mean local max and min, not global. In this case there are no global max or min values, because we can have [itex] f \rightarrow + \infty[/itex] by letting x go to zero from above (keeping y = 11-x), or letting x go to 11 from below (so y goes to zero from above). Similarly, we can have [itex] f \rightarrow -\infty[/itex] by letting x go to zero from below or go to 11 from above.

    RGV
     
    Last edited: Jun 18, 2012
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