I have the utility function:

$U(x_1,...,x_n)=a_0+\sum_{i=1}^{n}a_ix_i\;\;\;\;\;\;\;\;\;a_j\in\mathbb{R}_+ \;\;\forall j=\{0,...,n\}$ (maybe $a_0$ could be zero)

$\sum_{i=1}^{n}a_i\in (0,K)\;\;\;$ where $K<\infty$

Where the problem is:

$\max_{x_1,...,x_n}U\;\;\;\;\;\;\;\;c.t.\;\;\sum_{i=1}^{n}p_ix_i=m\;\;\;\;\;$ (With positive prices and income)

Which is the general case for perfect substitutes preferences for finding Walrasian-Marshallian demands. In the particular case $n=2$ the solution could be found geometrically, and the optimum turns out to be one of three cases (where two of them are corner solutions) depending on the relation between prices and parameters $a$. When $p_1/p_2>a_1/a_2$ solution is $(x_1^*=0, x_2^*=m/p_2)$, and the opposite case she spends all income in $x_1$, and the third case is a continuum of feasible points that satisfy the budget constraint.

So, here's the question: Is there a way to obtain a generalized Walrasian-Marshallian demand function (I imagine it would be by parts) for $n$-goods in this case, just like in Cobb-Douglas preferences that it turns out to be $x_i^m=\frac{a_im}{p_i\sum_{i=1}^{n}a_i}$ (please prove).


1 Answer 1


Let $b_i := \frac{a_i}{p_i}$, now choose the good that maximize $b_i$, let it be $b_j$ i.e. the $j$th good. Optimal choice is $x_j=m/p_j$, and $x_i = 0$ for $i\neq j$.

  • $\begingroup$ I get it! Do you mind telling me like a way to prove it? Thanks! $\endgroup$
    – manifold
    Jul 12, 2020 at 23:15
  • $\begingroup$ And other think, I get the intuition, but should not the parameter be compared as a weight versus all other parameters (like the $\frac{a_i}{\sum_{i=1}^{n}a_i}$ part in Cobb-Douglas)? Thanks! $\endgroup$
    – manifold
    Jul 12, 2020 at 23:23
  • $\begingroup$ Solve it for the 2 good case. Youll notice the pattern immediately. $\endgroup$
    – user28372
    Jul 13, 2020 at 3:45
  • 1
    $\begingroup$ (-1) Displaying formulas is not helping. $\endgroup$
    – Giskard
    Jul 13, 2020 at 8:08
  • 1
    $\begingroup$ @Giskard In my opinion, I have not displayed just the formula. The one who posted the question already knows how to solve the problem for the case of two goods, so according to me this is sufficient. $\endgroup$
    – Amit
    Apr 8, 2022 at 3:10

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