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Simple constraint problem bit confused

  1. Jan 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the maximum value of the function f(x, y) = x + y subject to the con-
    straint xy = 16, x > 0, y > 0.

    2. Relevant equations

    f(x, y) = x + y
    xy = 16

    3. The attempt at a solution

    I attempted this first using substitution, then using LaGrange multipliers. Both times I got an answer of 8. I entered it into Wolfram Alpha, got an answer of 8. But I'm convinced the answer should be 17 since f(16,1)=16+1=17 which also fits the constraint 16.1=16

    ???
     
  2. jcsd
  3. Jan 12, 2014 #2

    Dick

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    What you got is that x=y=4 is the only critical point. Getting a critical point doesn't automatically tell you it's a maximum. And I don't think 17 is what you are after either. 16=160*(1/10) as well.
     
  4. Jan 12, 2014 #3
    Thanks for the reply. I see your point but still don't know how to do the problem, anyway I'm going to bed now, I'll look at it tomorrow with fresh eyes.
     
  5. Jan 12, 2014 #4

    Dick

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    Sure, but my point was just that you might want to think about whether your problem even has a maximum.
     
  6. Jan 13, 2014 #5
    Well if 160*(1/10)=16 the value of the function is 160+(1/10), if 16000000*(1/1000000)=16 the value is 16000000+(1/10), so I can see that it tends toward infinity

    Still I'm not sure why the method of Lagrange doesn't work, and why does it give the answer (4,4). I know the gradients are in the direction at this point but I'm not sure what kind of point this is: Intuitively it's not a saddle point, or local max or local min.

    Finally what way should I answer this question in an exam situation?
     
  7. Jan 13, 2014 #6

    Dick

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    Why do you think (4,4) isn't a local min? In an exam I'd find the critical points and then give an argument like that to show there is no maximum.
     
  8. Jan 13, 2014 #7

    Ray Vickson

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    Basically, you are just facing a two-dimensional version of a familiar one-dimensional phenomenon. Take, for example, the simple function f(x) = x^2 in 1-d. If you set f'(x) = 0 you get x = 0 as the (only) stationary point. However, x = 0 is a minimum---the global minimum in this case---and there is NO maximum at all.

    The "theorems" about solving for max and or min by finding stationary points will (if stated properly) have hypotheses about the max and/or min actually existing in the region of interest. Just having a stationary point does not guarantee having an optimum, certainly not in a problem where there is no optimum. Example: f(x) = x^3 has no max and no min, but x = 0 is a stationary point.

    In your example, the point (x,y) = (4,4) is a (constrained) global min; there are second-order tests to ascertain this fact, but they tend to be a bit lengthy for use on a short class-quiz: they involve looking at the Hessian matrix of the Lagrangian in the tangent subspace of the constraint! However, in your case you can reduce the whole problem to a 1-d case by putting y = 16/x and then looking at max/min f(x) = (x + 16/x), x>0.
     
  9. Jan 15, 2014 #8
    Yeah, I just did it again... came out with x=y=λ=4. Doing the second-partials test would give an answer of 0 for D, meaning it's inconclusive. This is as far as the scope of my calculus course goes for these types of problems. I guess this is one they threw in just to see who the smart people were.
     
  10. Jan 15, 2014 #9

    Ray Vickson

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    Your second-order test is wrong: in a constrained problem you need to look at the Hessian matrix of the Lagrangian, not of the function you are maximizing or minimizing. In this case your Lagrangian is
    [tex] L = x+y + \lambda(xy-16). [/tex]
    The Hessian matrix of L is the matrix of second partial derivatives:
    [tex] H = \pmatrix{\frac{\partial^2 L}{\partial x^2} & \frac{\partial^2 L}{\partial x \partial y} \\
    \frac{\partial^2 L}{\partial y \partial x}& \frac{\partial^2 L}{\partial y^2}}
    = \pmatrix{0 & \lambda\\ \lambda & 0 } [/tex]
    We must project ##H## down into the tangent subspace of the constraint at the soluton point, and examine whether that lower-dimensional matrix is positive-definite, negative-definite, or indefinite; it turns out to be positive-definite, so the point (x,y) = (4,4) is a strict local minimum. As I already stated in my previous post, this is all a bit much for a quick class-quiz. However, the important lesson is that you were doing the wrong type of optimality test.

    You still have not dealt with the other issue several people have put to you: what about the maximum? Why does the Lagrange multiplier method fail to find it (if any)?
     
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