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I had this strange problem where the Lagrangian method gives the wrong answer in a constrained optimization problem. Here goes:

The problem is $$\max_{c,n,q} \alpha\log(c)+(1-\alpha)\log(nq)$$ subject to $$c+n+nq=1 \text{ and }c,n,q\geq 0$$ where $\alpha\in(0,1/2)$.

So I set up the Lagrangian: $$\mathcal{L}=\alpha\log(c)+(1-\alpha)\log(nq)+\lambda(1-c-n-nq)+\theta_c c+\theta_n n+\theta_q q.$$

The first order condition for $c$, $n$, and $q$ are as follows: $$\frac{\alpha}{c}+\theta_c=\lambda,$$ $$\frac{1-\alpha}{n}+\theta_n=\lambda(1+q),$$ and $$\frac{1-\alpha}{q}+\theta_q=\lambda n,$$ plus three Kuhn-Tucker conditions $$\theta_x x=0,x,\theta_x\geq 0$$ for $x=c,n,q$. Obviously, all three choice variables should be strictly positive. Hence the Lagrange multipliers $\theta_x=0$ for $x=c,n,q$. Then, from the last two FOCs it follows that $$\frac{q}{n}=\frac{1+q}{n},$$ which doesn't make sense at all. But I can't see what went wrong in the process. Please help me. Thanks very much!

oopsroger
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2 Answers2

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I believe the problem is that your inequality constraints are not truly $x\ge0$ but $x>0$.

When using Lagrange multipliers with inequality constraints to maximize a function, the assumption is that the inequality constraint can be written in the form $h(x)\le0$ and there two cases must be studied: maximization not on the border ($\theta_x=0$), and maximization on the border ($h(x)=0$, or in your case $x=0$). What your result is telling you is that there is no maximum unless it lies on the border (a maximum that might be a maximum only because there is a border look at the figure here).

By plotting $\alpha\log(x)+(1-\alpha)\log(y)$ in a 3D plot capable software for various values of $\alpha$ you probably will realize there is no maximum (I haven't done it).

What you should do is realize that since $c$, $n$, $q>0$, then your exact constraint implies $c$, $n$, $nq\le1$ and include that in your Langrange multipliers.

Kevin
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  • Also, the sign of your inequality terms is not right. – Kevin Jul 12 '14 at 16:31
  • thanks! Following your suggestion, I proved that there is no maximum for this problem. I put it in a separate answer (apparently it is easier to format). – oopsroger Jul 12 '14 at 16:57
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op here.

I got inspired by @kevin's suggestion that there is no maximum, and constructed the following proof.

For any combination of $(c,n,q)$ satisfying the constraint, we can construct $(c',n',q')$ as follows. $\forall k>0$, let $q'=10^k q$, $n'=10^{-k}n$. Since $n>0$, $n'<n$. Then we have $$c'=1-n'-n'q'=1-n'-nq>1-n-nq=c>0,$$ and it is clear that this new $(c',n',q')$ yields a higher functional value than $(c,n,q)$.

Hence the original optimization problem does not have a maximum.

oopsroger
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