# Consumer utility optimization question

I am having diffculty with this question: Can we proceed using Lagrange here since the utility function is convex, or do we have to treat them as perfect substitutes? If latter or any other method, please explain.

My attempt:

$$L= U(X,y) +λ(y-p1-p2)$$.

By using Lagrange, I got these FOCs: $$α Ax1^{(α -1)} - λp1=0$$& $$1-λp2=0$$.

So, is it correct to use Lagrange method here?

• Hi, Welcome to Economics SE! we have a policy regarding homework questions where we require that you show some work before an answer is provided. More on this topic here: economics.meta.stackexchange.com/questions/1465/… – EconJohn Sep 3 '17 at 21:38
• if you can put this into the body of your question it would make it more visual than looking at the comments. If you can also rescale and rotate your picture so people can see it more clearly it would be appreciated – EconJohn Sep 3 '17 at 21:49
• I personally am unable to read the question due to its orientation – EconJohn Sep 3 '17 at 21:50
• Done. Hope it helps now. – LUCIFER Sep 3 '17 at 21:58

## 1 Answer

I'll offer a little guidance to get you moving on the problem your own.

You've set up the Lagrangean incorrectly.

The maximization problem you have looks like:

$$\max \ U(x_1, x_2) \\ \text{s.t.} \ p_1 x_1 + p_2 x_2 \leq y$$

The Lagrangean should read:

$$\mathscr{L} = U(x_1, x_2) - \lambda (p_1 x_1 + p_2 x_2 - y)$$

The Lagrangean you offered had utility as a function of $X$ and $y$, but you can see your utility function does not have $y$ as an argument in it. So this is just a notation error to keep track of. As for the part with the Lagrangean multiplier, you can imagine that if the inequality constraint is binding, then $p_1 x_1 + p_2 x_2 - y = 0$, so in that case, maximizing the Lagrangean is just maximizing utility minus $0$, but it's a special $0$ that helps to incorporate the budget constraint into the problem. (As for the non-binding case, you can think about what is happening to the optimization problem yourself.)

Your first order conditions are correct though. So that tells me you've mostly made some typing errors. To answer your question about whether the Lagrangean is okay here, the Lagrangean will work as long as the optimum is an interior solution. So if you suspect either $x_1, x_2$ will be equal to zero, then you may have to break out Kuhn-Tucker's theorem. So think about under what conditions Lagrange's theorem holds.