I have the following utility function:


with budget constraint:

$$p_1x_1+p_2x_2+p_3x_3\leq I$$

I use the Kuhn-Tucker method to find the optimal choices of the Utility maximization problem. My equations are:

$$x_2-\lambda p_1+M_1=0$$ $$x_1-\lambda p_2+M_2=0$$

$$1-\lambda p_3+M_3=0$$

$$p_1M_1=0$$ $$p_2M_2=0$$ $$p_3M_3=0$$ $$p_1x_1+p_2x_2+p_3x_3-I=0$$

$$(M_1,M_2,M_3,\lambda \geq 0)$$

When I set $M_1=M_2=M_3=0$ (Lagrangian case), I got the optimal solutions for $x_1,x_2$ as:

$$x_1=\frac{p2}{p3}$$ and $$x_2=\frac{p_1}{p_3}$$

How could I construct a Marshallian demand function in this case? The optimal solutions haven't got the I (income) variable.

Is it correct to define a Marshallian demand function for good $x_1$ as: $x_1(p,I)=\frac{p_2}{p_3}$?

Marshallian demand function (named after Alfred Marshall) specifies what the consumer would buy in each price and income or wealth situation, assuming it perfectly solves the utility maximization problem


3 Answers 3


These are proper Marshallian demand functions, even though Income does not appear in them. This is due to specific form of the utility function (and the candidate solution of all goods being purchased at strictly positive quantities). It emerges that there is no income effect for goods $x_1$ and $x_2$ - optimal uncompensated demand does not depend after all on the level of income, but only on the relative prices.

The reason why Marshallian demand is defined as it is, is to make clear that it does not include any kind of "income compensation" as Hicksian demand does. But this does not preclude a case like yours, which again, depends on the form of the utility function.

Thinking economically, what goods can you think of whose demand may realistically not depend on the level of income but only on relative prices? Answering this question would be useful in order to map the mathematical expression of the utility function to real-world economic phenomena.

Also, note that the specific utility function points to "quasi-linear" frameworks, where the good that enters additively, essentially functions as Income itself.

  • $\begingroup$ $x_1=\frac{p_2}{p_3}$ and $x_2=\frac{p_1}{p_3}$ cannot be the demand. To see this consider, $p_1=p_2=4$, $p_3=1$ and $I=40$. Given these expressions, consumer should demand $(4,4,8)$, but $u(4,4,8)=24<40=u(0,0,40)$. $\endgroup$
    – Amit
    Commented Oct 16, 2023 at 0:24
  • $\begingroup$ @Amit Correct, as regards the overall view of the problem. But note that my answer was a response to a specific narrow question of the OP, who was examining the case of all goods being strictly positive. I am looking now at your full treatment of the problem. $\endgroup$ Commented Oct 16, 2023 at 14:28

Given that $u(x_1,x_2, x_3)=x_1x_2+x_3$, demand is the solution to the following problem: \begin{eqnarray*} \max_{x_1,x_2,x_3} & x_1x_2+x_3 \\ \text{s.t. } & p_1x_1+p_2x_2+p_3x_3 \leq I \\ \text{and } & x_1\geq 0, \ x_2\geq 0, \ x_3\geq 0\end{eqnarray*} where $p_1>0$, $p_2>0$, $p_3>0$, and $I\geq 0$ are given. Please note that the objective is not quasi-concave. To see this, consider $(x_1', x_2', x_3')=(2,2,0)$, $(x_1'', x_2'', x_3'')=(0,0,4)$, and $\lambda=\frac{1}{2}$. Clearly,

$u(\lambda (x_1', x_2', x_3')+(1-\lambda)(x_1'', x_2'', x_3''))=u(1,1,2)=3<4=\min(u(x_1', x_2', x_3'),u(x_1'', x_2'', x_3''))$

Solving for demand we get \begin{eqnarray*} (x_1^d,x_2^d,x_3^d)(p_1,p_2,p_3,I)\in \begin{cases}\left\{\left(\dfrac{I}{2p_1},\dfrac{I}{2p_2},0\right)\right\} &\text{if } \dfrac{I^2}{4p_1p_2}>\dfrac{I}{p_3} \\ \left\{\left(0,0,\dfrac{I}{p_3}\right)\right\} &\text{if } \dfrac{I^2}{4p_1p_2}<\dfrac{I}{p_3} \\ \left\{\left(\dfrac{I}{2p_1},\dfrac{I}{2p_2},0\right),\left(0,0,\dfrac{I}{p_3}\right)\right\} &\text{if } \dfrac{I^2}{4p_1p_2}=\dfrac{I}{p_3} \end{cases}\end{eqnarray*}

Note: Approach I have used to find the demand here is same as the one I have used here: https://economics.stackexchange.com/a/51562/11824


In response to a correct comment left by user @Amit below my 8-years-old answer, I am adding another answer here to provide intuition and some simple analytical steps that support user's @Amit full but compactly written solution to this utility maximization problem.

The moral here will turn out to be, indeed, that it is never optimal to consume all three goods at strictly positive quantities.

Our Lagrangian is $$\Lambda = x_1x_2 + x_3 +\lambda(I-p_1x_1 - p_2x_2 -p_3x_3) + \mu_1x_1 + \mu_2x_2 + \mu_3 x_3 . $$

But we will not need to perform the obvious computations of partial derivatives and the like.

First, one must realize that it is never optimal to consume $x_1$ but not $x_2$, or $x_2$ but not $x_1$, because in both case the utility derived from the consumed good will be zero. This leaves us with three scenarios: $$[A]:\quad x_1, x_2, x_3 > 0\\ [B]:\quad x_1, x_2 > 0, \quad x_3 =0\\ [C]:\quad x_1=x_2=0, \quad x_3 > 0. $$

SCENARIO [B] $x_1, x_2 > 0, \quad x_3 =0$

Here, maximized utility will be $$U^*_B = x_1^*x_2^*,\qquad s.t.\quad p_1x_1^* + p_2x_2^* = I.$$

If one uses the constraint to substitute back in the maximized utility function (say, for $x_1^*$) and maximize w.r.t. $x_2^*$, we will get $$x_1^* = \frac{I}{2p_1},\; x_2^* = \frac{I}{2p_2} \implies U^*_B = \frac{I^2}{4p_1p_2}.$$

SCENARIO [C] $x_1=x_2=0, \quad x_3 > 0$

Here, maximized utility will be $U^*_C = x_3^*$, and it is immediate that $$x_3^*= \frac{I}{p_3}\implies U^*_C = \frac{I}{p_3}.$$

From the above results comes the solution provided by user @Amit: it emerges that depending on the level of income in relation to prices, if $(I^2/4p_1p_2) > I/p_3$ it is better to not consume $x_3$ at all, etc.

Note that the threshold value of the level of income simplifies to $$I >< \frac{4p_1p_2}{p_3}.$$

Keep that.

SCENARIO [A] $\quad x_1, x_2, x_3 > 0.$

But let's also consider the case that the OP originally considered, i.e. where we consume all three goods in strictly positive quantities. Here the solutions are those found by the OP, namely,

$$x_1^* = \frac{p_2}{p_3},\; x_2^* = \frac{p_1}{p_3}$$ subject to $$p_1x_1^* + p_2x_2^* + p_3x_3^* = I \implies x_3^* = \frac{I}{p_3} - \frac{2p_1p_2}{p_3^2}.$$

From this to be feasible (i.e. have $x_3^* > 0$) it must be the case that

$$\frac{I}{p_3} - \frac{2p_1p_2}{p_3^2} > 0 \implies I > \frac{2p_1p_2}{p_3}.$$

Combining this with the results we obtained earlier comparing scenaria $[B]$ and $[C]$ we have two additional situations to consider

SCENARIO $[A,C]: \quad \frac{2p_1p_2}{p_3} < I < \frac{4p_1p_2}{p_3}.$

The left inequality says that it is feasible to have all three goods positive, the right equality says it is preferable to consume just $x_3$ compared to consume just $(x_1, x_2)$. But here we want to compare consuming $x_3$ only against consuming all three $(x_1, x_2, x_3)$.

What is the utility under scenario $[A]$? It is

$$U^*_A = \frac{p_1p_2}{p_3^2} + x_3^* = \frac{p_1p_2}{p_3^2} + \frac{I}{p_3} - \frac{2p_1p_2}{p_3^2} = \frac{I}{p_3} - \frac{p_1p_2}{p_3^2} < \frac{I}{p_3} = U^*_C.$$

So in this intermediate range of income, it is preferable to consume only $x_3$.

SCENARIO $[A,B]: \quad \frac{2p_1p_2}{p_3} < \frac{4p_1p_2}{p_3}< I$

Here we compare the prospect of consuming all three goods, against consuming only $(x_1, x_2)$. We have as before

$$U^*_A = \frac{I}{p_3} - \frac{p_1p_2}{p_3^2}$$

to be compared against

$$U^*_B = \frac{I^2}{4p_1p_2}.$$

Form the polynomial $$P_{B,A} \equiv U^*_B - U^*_A = \frac{I^2}{4p_1p_2} - \frac{I}{p_3} + \frac{p_1p_2}{p_3^2}.$$

Its discriminant is

$$\Delta = \left(\frac{1}{p_3}\right)^2 - 4\frac{1}{4p_1p_2}\frac{p_1p_2}{p_3^2} = 0,$$

so the polynomial has a double root which is $$I_{d}= \frac{1/p_3}{2/4p_1p_2} = \frac{2p_1p_2}{p_3},$$

and the polynomial can be written $$P_{B,A} \equiv U^*_B - U^*_A =\frac{1}{4p_1p_2}\left(I- \frac{2p_1p_2}{p_3}\right)^2 > 0.$$

The inequality is strict because in the case we are considering, income $I$ is strictly greater than $2p_1p_2 / p_3$. So we conclude that, comparing consuming $(x_1, x_2, x_3)$ against just $(x_1, x_2)$, we prefer the latter.

In all therefore, it is never optimal to consume all three goods: depending on the income level, we should either consume just $(x_1, x_2)$, or just $(x_3)$.


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