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How to visualize Constraints

  1. Mar 27, 2012 #1
    That’s an issue for me. I don’t know how should I visualize constraints in constrained optimization problems in R^{3}. This is a problem cause I cannot see how it works the multiplier solution in the inequality constraints. That’s not a matter of exercises, that’s really a problem of visualization, nothing more.

    Btw, the next one is not an homework…LOL, it’s just a way to explain my problem with a function that it’s easy to imagine.

    Let’s say that I have to face a constrained maximization problem and the function is something quite common like a Cobb-Douglas function with the parameter set to .5.
    Now, let’s add a constraint like x[itex]^{2}[/itex] + y[itex]^{2}[/itex] ≤1.

    What is this?
    I mean, how do I have to consider it in terms of visualization of the problem in the space. Is it simply a set defined on [itex]\Re[/itex][itex]^{2}[/itex] that creates a constraint on [itex]\Re[/itex][itex]^{2}[/itex]?
    However, if this is the case what’s the point of talking about the gradient of this constraint function that has to go in the same direction of the object function? There is no such a thing if we are dealing only with something that is on [itex]\Re[/itex][itex]^{2}[/itex].

    Or do I have to consider it a function, specifically g(x,y)=x[itex]^{2}[/itex] + y[itex]^{2}[/itex]? If this is the case, what the heck is this? A paraboloid that stops at a certain point?

    I am really lost…

    Thanks in advance.

    PS: The problem is the same with linear constraints. Are them lines or planes?

    PPS: Sorry for all these easy (borderline stupid) questions, but that’s the problem when you have to figure out everything by yourself and there is nobody around that can explain you…that’s why I think this forum is amazing. :smile:
     
  2. jcsd
  3. Mar 27, 2012 #2

    Stephen Tashi

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    Science Advisor

    You ought to not to confuse the concepts of a "constraint" versus "the set of points that satisfy a constraint". It is straightforward to test whether a point [itex] (x,y,z) [/itex] satisfies the constraint [itex] x^2 + y^2 \leq 1 [/itex]. The set of points in 3 dimensions that do form a cylindrical solid - or I should say an infinitely long cylindrical sold.
     
    Last edited: Mar 27, 2012
  4. Mar 28, 2012 #3
    Ok, which means that if we move to linear constraints we have that each linear constraint is an infinite long plane, that is orthogonal to [itex]\Re^{2}[/itex].
    Am I right?

    If this is the case there is something that is still not clear to me.
    Stylized fact: the gradient of this infinitely long cylindrical solid has to point in the same direction of the gradient of the objective function.
    When we analyze equality constraints, the point that books emphasize is that it is not relevant if the gradient of the constraint goes in the same direction of the gradient of the objective function. Fair enough from an algebraic point of view, but in terms of visualization why should we consider the possibility that the gradient doesn't go in the same direction of the gradient of the objective function?
    What's the difference in terms of visualization between this case and the inequality one (between the cylindrical solid orthogonal to [itex]\Re^{2}[/itex] and the plane)?

    Honestly, the standard pic used in the presentation of the (in-)equality constraint problem, with level curves of the objective function and arrows related to the gradients of the objective function and the constraints, is a bit problematic for me.
     
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