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Minima of a multivariable function.

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the minimum value of [tex]f(x,y)=e^{x+y}-2[/tex] within x≥0 and y≥0.

    2. Relevant equations
    [tex]D=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2[/tex]
    Answer is -1

    3. The attempt at a solution
    So for all partial derivatives I got [tex]e^{x+y}[/tex] (and mixed), but when I calculate the discriminate (subbing in (0,0) I get [tex]D=e⋅e-(e)^2=0[/tex].

    I was just confused on how the answer is -1 when the discriminate has to be larger then zero to be a minimum point.
     
  2. jcsd
  3. May 3, 2015 #2

    ehild

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    Does the function have local extrema? What is the condition that a local extreme exist?
     
  4. May 3, 2015 #3

    Ray Vickson

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    When you have constraints (even simple ones like x,y >= 0) the usual first-and second-order tests must be modified. In other words, you cannot just look at the gradient and Hessian of f(x,y) at the solution point, if that point lies on the boundary. In such a problem the discriminant does NOT have to be > 0.

    Anyway, in problems where it applies that is just a sufficient condition, not a necessary one. A correct statement---but not in this problem---is that if the discriminant is > 0 the point is a strict local minimum. The corresponding necessary condition is that the discriminant be >= 0, (not > 0).

    In your case, the function f(x,y) is monotonically strictly increasing in the region x >=0, y >= 0, so the solution is at the boundary x = y = 0. Basically, no fancy tests are needed.
     
  5. May 3, 2015 #4
    So it's simply [tex]e^{(0)(0)}-2=-1?[/tex]
     
  6. May 4, 2015 #5

    HallsofIvy

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    Yes. Since the partial derivatives of [itex]e^{x+y}-2[/itex] are both positive for all [itex]x\ge 0[/itex], [itex]y\ge 0[/itex] are positive this function is always increasing as x and y increase so its minimum value must occur at (0, 0).
     
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