This problem I am working on comes out of--surprise--the Mas-Colell book for graduate micro (3.D.6). I think I have correctly used the FOC of the Lagrangian of the utility maximization problem to derive the consumer's Walrasian demand. My answer does not match the book. We are given that

$u(x) = (x_1-b_1)^\alpha (x_2-b_2)^\beta(x_3-b_3)^\gamma$

(and then from the first part we can say that $\alpha + \beta + \gamma = 1 $ WLOG.)

The solutions say that we take a monotonic transformation:

$\ln(u(x)) = \alpha \ln(x_1-b_1) + \beta \ln(x_2-b_2) + \gamma \ln(x_3-b_3)$

and then I set up the Lagrangian of this:

$ \mathcal{L} = \alpha \ln(x_1-b_1) + \beta \ln(x_2-b_2) + \gamma \ln(x_3-b_3) - \lambda(p_1x_1 + p_2x_2 + p_3x_3 - w)$

and the FOCs are:

$\frac{\alpha}{x_1-b_1} - \lambda p_1 = 0$

$\frac{\beta}{x_2-b_2} - \lambda p_2 = 0$

$\frac{\gamma}{x_3-b_3} - \lambda p_3 = 0$

Solve for the x's:

$x_1 = \frac{\alpha}{\lambda p_1} + b_1$

$x_2 = \frac{\beta}{\lambda p_2} + b_2$

$x_3 = \frac{\gamma}{\lambda p_3} + b_3$

Which leads us to:

$$x(p,w) = (b_1, b_2, b_3) + \left(\frac{\alpha}{\lambda p_1},\frac{\beta}{\lambda p_2},\frac{\gamma}{\lambda p_3}\right)$$

This is not what the book got, so I used Walras law to get the desired result:

$p \cdot x = w$

$\implies w - (b \cdot p) = \frac{1}{\lambda}(\alpha + \beta + \gamma)= \frac{1}{\lambda}$

$$\implies x(p,w) = (b_1, b_2, b_3) + (w - (b \cdot p))\left(\frac{\alpha}{p_1},\frac{\beta}{p_2},\frac{\gamma}{p_3}\right)$$

which is the book's solution.

So my questions are:

Did I do the derivation right? How should I know to take a log transformation of the original utility function? Is there any particular information that is supposed to tip me off?

  • 1
    $\begingroup$ Rhetorically, one would first wonder how the setup is preventing $w<b_1 p_1+b_2 p_2+b_3 p_3$. Note: it is not in the book. $\endgroup$
    – ramazan
    Commented Oct 10, 2015 at 3:28
  • $\begingroup$ The problem is I am not exactly sure on what the intuition is behind what $b_1, b_2, b_3$ stand for in the function, if they have some economic interpretation. $\endgroup$
    – Kitsune Cavalry
    Commented Oct 10, 2015 at 3:38

1 Answer 1


I am not sure I understand your question. $\lambda$ is a Lagrange multiplicator which has a value in the optimum. It is not a parameter and hence you cannot leave it in your solution.
When solving the Lagrangian the optimal solution has the form $(x,\lambda)$ and in addition to your conditions $$x_1 = \frac{\alpha}{\lambda p_1} + b_1$$ $$x_2 = \frac{\beta}{\lambda p_2} + b_2$$ $$x_3 = \frac{\gamma}{\lambda p_3} + b_3$$ it also has to fulfill $$ (p \cdot x - w) \lambda = 0. $$ If $\lambda \neq 0$ this also means $p \cdot x = w$. $\lambda$ cannot be equal to zero because it is in the denominator of your optimal solution for $x$: $$x(p,w) = (b_1, b_2, b_3) + \left(\frac{\alpha}{\lambda p_1},\frac{\beta}{\lambda p_2},\frac{\gamma}{\lambda p_3}\right)$$ and hence you were correct to use the Walras law or budget constraint.

For more on optimality conditions see the Karush-Kuhn-Tucker theorem.

About $b_1, b_2$ and $b_3$: I think this is supposed to be a bliss point.

  • $\begingroup$ My question is mainly why am I supposed to take a monotonic transformation of the original utility function before taking the Lagrangian. I understand that it makes my life easier and still represents u, but why log specifically? $\endgroup$
    – Kitsune Cavalry
    Commented Oct 10, 2015 at 6:00
  • 1
    $\begingroup$ Ah, I see. Sorry, I missed that. Yes I think that just makes your life easier because this way you get exactly one $x$ variable in every optimaility condition. I think this log transform is frequently useful but not always. $\endgroup$
    – Giskard
    Commented Oct 10, 2015 at 6:03
  • $\begingroup$ Okay, that seems pretty sensible in hindsight. Log is such a nice function sometimes. $\endgroup$
    – Kitsune Cavalry
    Commented Oct 10, 2015 at 6:08

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