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I have the following utility maximization problem: $$\max (xy)$$ $$(x+y-2)^2 \leq 0$$ Conditions: $$y-2\lambda (x+y-2) =0$$ $$x-2\lambda (x+y-2) =0$$ $$\lambda(x+y-2)^2=0$$

When I set $\lambda>0$, I get: $$(x+y-2)^2=0 \Rightarrow (x+y-2) = 0$$ $$y-2\lambda (x+y-2) = y = 0$$ $$x-2\lambda (x+y-2) = x = 0$$

But the obvious solution is $x=y=1$.

When I set $$\lambda=0$$ that's not a case with a valid solution either. Is the constraint too low? What is the explanation to this?

I would greatly appreciate your help.

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  • $\begingroup$ I tried to simplify the equations a bit. If you disagree feel free to roll back my edit. $\endgroup$ – Giskard Oct 20 '15 at 7:04
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As @user32416 pointed out the first order stationarity conditions are not enough. Specifically it seems that you violate Slater's condition, which states that "the feasible region must have an interior point". There are no $x,y$ for which $$(x+y-2)^2 < 0.$$

If you rephrase the problem to $$\max (xy)$$ $$x+y-2 = 0$$ $$x,y \geq 0$$ Slater's condition is met (for linear conditions no interior points are necessary) and you can apply Karush-Kuhn-Tucker.

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This is an ill-posed question. Even without going through KKT, your constraint $(x + y - 2)^2 \le 0$, since the left-hand side is a square, means that the only solution that is feasible is the one where the equality binds; i.e. $(x + y - 2)^2 = 0$, or that $|x + y - 2| = 0$ -- of which what you say that $x = y = 1$ is a solution.

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  • $\begingroup$ Yes but can you apply K-K-T to this rephrased problem? $\endgroup$ – Giskard Oct 20 '15 at 6:55
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    $\begingroup$ You're only including the first order stationarity conditions here. But you're completely ignoring the complementary slackness and primal feasibility conditions. Thus what you'd written there is not even the full KKT formulation. $\endgroup$ – user32416 Oct 20 '15 at 7:01
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    $\begingroup$ Thank you all for your constructive criticism and help! My prof said that this problem wasn't well defined, because he wanted to represent the importance of Slater's condition. $\endgroup$ – Übel Yildmar Oct 20 '15 at 22:00
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As already noted, the constraint is such that it turns itself into an equality constraint. Your problem then is simplified to

$$\max_{x,y} (xy) \;\;\; s.t.\;\; x+y-2=0$$

Rearrange the constraint and substitute for $y$ to obtain the single variable unconstrained problem

$$\max_{x} (2x-x^2)$$

This provides immediately the solution $x=1$ and so also $y=1$.

Using the Karush-Kuhn-Tucker conditions on the original problem, may be good practice in order to see for yourself that the complementary slackness condition must also hold (and "Slater's condition" is one of the formulations of it), but Occam's razor would require that the problem should be actually solved as above.

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