I am having difficulty with a particular bliss point problem. The basic issue I have is my approaches seem flawed and I can't tell why.

The equation is

$$U(x,y) = 36x -4x^2 + 6y-2y^2$$

subject to

$$ 5x + 7y=40$$

My first instinct was to use the method of Lagrangian multipliers. But I had this on an exam this afternoon and my professor said I shouldn't use it here. I don't understand why. I tried it using wolfram alpha and also solved it by hand and got the same results


approximately as the optimum bundle. But seemingly this isn't the train of thought I was supposed to use.

I then had another idea. I can solve for $$U_x=U_y=0$$ because at the bliss point you should have MU equal to zero.

\begin{align*} 36-8x &= 0 \\ x&=\frac{36}{8}=4.5\\ 6-4y&=0\\ y&=\frac{6}{4}=1.5 \end{align*}

But this lead to different results than just using TMoLM.

My Question:

Can someone explain why TMoLM isn't the right approach? Why does it lead to different results then what my professor suggested and what others have said in the comments?

  • $\begingroup$ Your second idea is good, but you made a typo: $$ y = \frac{6}{4} = 1.5. $$ $\endgroup$ – Giskard May 8 '15 at 6:47
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    $\begingroup$ Looking at it a different way might help you. $$ U(x,y) = 36\cdot x−4\cdot x^2+6\cdot y−2\cdot y^2 = (2\cdot x - 9)^2 + 2\cdot(y - 1.5)^2 + constant $$ For any utility level $\bar{u}$ the equation $U(x,y) = \bar{u}$ characterizes a shape that should be familiar to you (if you know coordinate geometry). If you know the shape of the level curves finding the optimum or the optimum condition should be easy. $\endgroup$ – Giskard May 8 '15 at 9:13
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    $\begingroup$ @denesp I think you should post that as an answer. $\endgroup$ – FooBar May 8 '15 at 13:11
  • $\begingroup$ I agree, that is an answer and I would upvote....but...I would like an explanation why TMoLM isn't applicable there. That is bothering me. It should be usable. And it doesn't yield the same results. That is really why I asked this question in the first place. $\endgroup$ – Stan Shunpike May 8 '15 at 15:01

It all depends on whether you treat the budget constraint as an equality or inequality constraint. These are two different problems, with two different solutions in this case.

One version of the problem (rewriting the objective in the form suggested by denesp, and dropping the constant, for clarity) is \begin{align} \max~&-4(x-4.5)^2 -2(y-1.5)^2\\\text{s.t. }&5x+7y=40 \end{align} With this equality constraint, the optimum is $(x,y)=(4.78,2.29)$ as you found using the method of Lagrange multipliers and Wolfram alpha. It's true that you would prefer consuming less and staying at the bliss point, but if we assume that the budget constraint is an equality then you aren't allowed to consume less.

Another version of the problem is \begin{align} \max~&-4(x-4.5)^2 -2(y-1.5)^2\\\text{s.t. }&5x+7y\leq40 \end{align} with an inequality in the budget constraint. Here, as you point out, it's clear that the optimum is $(x,y)=(4.5,1.5)$, because that's the consumption profile that globally maximizes the utility function (and it obeys the budget inequality).

Assuming that the budget constraint is an inequality is usually more appropriate (presuming that you can always "throw away" extra wealth).

When solving constrained optimization problems where some of the constraints are inequalities, we need a generalization of the basic method of Lagrange multipliers called the Karesh-Kuhn-Tucker (KKT) conditions, which has probably not been covered in your course. You can read about this in more detail in the linked wikipedia article, but the basic idea is simple: we can still set up the Lagrangian, where we write the constraint as $g(x,y)=5x+7x-40\leq 0$ and then subtract $\lambda g(x,y)$ from the objective: $$\mathcal{L} = -4(x-4.5)^2 -2(y-1.5)^2 - \lambda(5x+7y-40)$$ We can then equate the partial derivatives to 0 ($\partial\mathcal{L}/\partial x = 0$ and $\partial\mathcal{L}/\partial y=0$) to get two conditions \begin{align} \frac{\partial\mathcal{L}}{\partial x}&=0 \Longleftrightarrow 8(x-4.5)=-5\lambda\tag{1}\\ \frac{\partial\mathcal{L}}{\partial y}&=0 \Longleftrightarrow 4(y-1.5)=-7\lambda\tag{2} \end{align} The question, then, is how to pin down $\lambda$. You're probably used to finding $\lambda$ by seeing which value causes the budget constraint to hold with equality. In this case, however, that implies $\lambda<0$. This is not allowed for an inequality constraint under the KKT conditions, which require $\lambda\geq 0$.** Since $\lambda$ turns out to be the marginal utility of wealth, this constraint is tantamount to saying that more wealth (which relaxes the inequality) shouldn't hurt us.

The KKT conditions also have a requirement called complementary slackness, which states that either the inequality must bind (holding with equality) at the optimum, or else $\lambda$ must be 0. (Since $\lambda$ measures the "cost" of the constraint, this is logical: if the constraint isn't binding, then the cost must be 0, and it shouldn't be affecting your local optimization problem.) We already saw that if the budget constraint holds with equality, we get $\lambda<0$, which isn't allowed, so complementary slackness implies our only remaining option is $\lambda=0$. Plugging this into (1) and (2) above, we get $(x,y) = (4.5,1.5)$, which is the solution you already found intuitively.

Usually, there is no need in introductory micro to use the full KKT conditions, because we work with utility functions that are monotonic, where you'll always want to exhaust your budget constraint. Simple concave quadratic utility with an interior bliss point, like in this case, is an exception where we need a very simple case of the KKT conditions to do the optimization properly.

**Note: of course, if we add $\lambda g(x)$ rather than subtract it to the Lagrangian, the signs reverse. I've seen many different conventions, and I'm going with the one that makes $\lambda$ nonnegative. The Wikipedia article uses the opposite convention.

  • $\begingroup$ +1 my professor has mentioned KKT conditions as well as complementary slackness. But I'm not sure I follow. In particular the "pin down the lambda" portion. Why does the sign matter? $\endgroup$ – Stan Shunpike May 9 '15 at 4:36
  • $\begingroup$ Does $L$ vs $\mathcal{L}$ mean something different? $\endgroup$ – Stan Shunpike May 11 '15 at 3:10
  • $\begingroup$ sorry, "pin down" is just my way of saying "solve for". the sign of $\lambda$matters because when the constraint is an inequality, only one sign makes sense: you can interpret $\lambda$ as the cost of the constraint (or equivalently the value you'd get from relaxing it), and you can't have a negative cost, because then you'd want to move away from the constraint. $\endgroup$ – nominally rigid May 12 '15 at 3:24
  • $\begingroup$ ($\mathcal{L}$ is just my notation for the Lagrangian - don't know how common or appropriate this is.) $\endgroup$ – nominally rigid May 12 '15 at 3:25
  • $\begingroup$ Okay. In physics, they mean very different things. That's why I asked. $\endgroup$ – Stan Shunpike May 12 '15 at 3:28

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