<p>Given that <span class="math-container">$u(x_1,x_2, x_3)=x_1x_2+x_3$</span>, demand is the solution to the following problem:
<span class="math-container">\begin{eqnarray*} \max_{x_1,x_2,x_3} &amp; x_1x_2+x_3 \\ \text{s.t. } &amp; p_1x_1+p_2x_2+p_3x_3 \leq I \\ \text{and } &amp; x_1\geq 0, \ x_2\geq 0, \ x_3\geq 0\end{eqnarray*}</span>
where <span class="math-container">$p_1&gt;0$</span>, <span class="math-container">$p_2&gt;0$</span>, <span class="math-container">$p_3&gt;0$</span>, and <span class="math-container">$I\geq 0$</span> are given.
Please note that the objective is not quasi-concave. To see this, consider <span class="math-container">$(x_1', x_2', x_3')=(2,2,0)$</span>, <span class="math-container">$(x_1'', x_2'', x_3'')=(0,0,4)$</span>, and <span class="math-container">$\lambda=\frac{1}{2}$</span>. Clearly,</p>
<p><span class="math-container">$u(\lambda (x_1', x_2', x_3')+(1-\lambda)(x_1'', x_2'', x_3''))=u(1,1,2)=3&lt;4=\min(u(x_1', x_2', x_3'),u(x_1'', x_2'', x_3''))$</span></p>
<p>Solving for demand we get
<span class="math-container">\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\} &amp;\text{if } \dfrac{I^2}{4p_1p_2}&gt;\dfrac{I}{p_3} \\ \left\{\left(0,0,\dfrac{I}{p_3}\right)\right\} &amp;\text{if } \dfrac{I^2}{4p_1p_2}&lt;\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\} &amp;\text{if } \dfrac{I^2}{4p_1p_2}=\dfrac{I}{p_3} \end{cases}\end{eqnarray*}</span></p>
<p>Note: Approach I have used to find the demand here is same as the one I have used here: <a href="https://economics.stackexchange.com/a/51562/11824">https://economics.stackexchange.com/a/51562/11824</a></p>

CC BY-SA 4.0
2023-10-15T23:55:00.323Z
2023-10-16T00:10:11.440Z
2023-10-16T00:10:11.440Z
Amit
<p>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.</p>
<p>The moral here will turn out to be, indeed, that it is never optimal to consume all three goods at strictly positive quantities.</p>
<p>Our Lagrangian is
<span class="math-container">$$\Lambda = x_1x_2 + x_3 +\lambda(I-p_1x_1 - p_2x_2 -p_3x_3) + \mu_1x_1 + \mu_2x_2 + \mu_3 x_3 . $$</span></p>
<p>But we will not need to perform the obvious computations of partial derivatives and the like.</p>
<p>First, one must realize that it is never optimal to consume <span class="math-container">$x_1$</span> but not <span class="math-container">$x_2$</span>, or <span class="math-container">$x_2$</span> but not <span class="math-container">$x_1$</span>, because in both case the utility derived from the consumed good will be zero. This leaves us with three scenarios:
<span class="math-container">$$[A]:\quad x_1, x_2, x_3 &gt; 0\\
[B]:\quad x_1, x_2 &gt; 0, \quad x_3 =0\\
[C]:\quad x_1=x_2=0, \quad x_3 &gt; 0.
$$</span></p>
<p><strong>SCENARIO [B] <span class="math-container">$x_1, x_2 &gt; 0, \quad x_3 =0$</span></strong></p>
<p>Here, maximized utility will be
<span class="math-container">$$U^*_B = x_1^*x_2^*,\qquad s.t.\quad p_1x_1^* + p_2x_2^* = I.$$</span></p>
<p>If one uses the constraint to substitute back in the maximized utility function (say, for <span class="math-container">$x_1^*$</span>) and maximize w.r.t. <span class="math-container">$x_2^*$</span>, we will get
<span class="math-container">$$x_1^* = \frac{I}{2p_1},\; x_2^* = \frac{I}{2p_2} \implies U^*_B = \frac{I^2}{4p_1p_2}.$$</span></p>
<p><strong>SCENARIO [C] <span class="math-container">$x_1=x_2=0, \quad x_3 &gt; 0$</span></strong></p>
<p>Here, maximized utility will be <span class="math-container">$U^*_C = x_3^*$</span>, and it is immediate that
<span class="math-container">$$x_3^*= \frac{I}{p_3}\implies U^*_C = \frac{I}{p_3}.$$</span></p>
<p>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 <span class="math-container">$(I^2/4p_1p_2) &gt; I/p_3$</span> it is better to not consume <span class="math-container">$x_3$</span> at all, etc.</p>
<p>Note that the threshold value of the level of income simplifies to
<span class="math-container">$$I &gt;&lt; \frac{4p_1p_2}{p_3}.$$</span></p>
<p>Keep that.</p>
<hr />
<p><strong>SCENARIO [A]</strong> <span class="math-container">$\quad x_1, x_2, x_3 &gt; 0.$</span></p>
<p>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,</p>
<p><span class="math-container">$$x_1^* = \frac{p_2}{p_3},\; x_2^* = \frac{p_1}{p_3}$$</span>
subject to
<span class="math-container">$$p_1x_1^* + p_2x_2^* + p_3x_3^* = I \implies x_3^* = \frac{I}{p_3} - \frac{2p_1p_2}{p_3^2}.$$</span></p>
<p>From this to be feasible (i.e. have <span class="math-container">$x_3^* &gt; 0$</span>) it must be the case that</p>
<p><span class="math-container">$$\frac{I}{p_3} - \frac{2p_1p_2}{p_3^2} &gt; 0 \implies I &gt; \frac{2p_1p_2}{p_3}.$$</span></p>
<p>Combining this with the results we obtained earlier comparing scenaria <span class="math-container">$[B]$</span> and <span class="math-container">$[C]$</span> we have two additional situations to consider</p>
<p><strong>SCENARIO</strong> <span class="math-container">$[A,C]: \quad \frac{2p_1p_2}{p_3} &lt; I &lt; \frac{4p_1p_2}{p_3}.$</span></p>
<p>The left inequality says that it is feasible to have all three goods positive, the right equality says it is preferable to consume just <span class="math-container">$x_3$</span> <em>compared to consume just</em> <span class="math-container">$(x_1, x_2)$</span>. But here we want to compare consuming <span class="math-container">$x_3$</span> only against consuming all three <span class="math-container">$(x_1, x_2, x_3)$</span>.</p>
<p>What is the utility under scenario <span class="math-container">$[A]$</span>? It is</p>
<p><span class="math-container">$$U^*_A = \frac{p_1p_2}{p_3^2} + x_3^* = \frac{p_1p_2}{p_3^2} + \frac{I}{p_3} - \frac{2p_1p_2}{p_3^2} = \frac{I}{p_3} - \frac{p_1p_2}{p_3^2} &lt; \frac{I}{p_3} = U^*_C.$$</span></p>
<p>So in this intermediate range of income, it is preferable to consume only <span class="math-container">$x_3$</span>.</p>
<p><strong>SCENARIO</strong> <span class="math-container">$[A,B]: \quad \frac{2p_1p_2}{p_3} &lt;  \frac{4p_1p_2}{p_3}&lt; I$</span></p>
<p>Here we compare the prospect of consuming all three goods, against consuming only <span class="math-container">$(x_1, x_2)$</span>. We have as before</p>
<p><span class="math-container">$$U^*_A = \frac{I}{p_3} - \frac{p_1p_2}{p_3^2}$$</span></p>
<p>to be compared against</p>
<p><span class="math-container">$$U^*_B = \frac{I^2}{4p_1p_2}.$$</span></p>
<p>Form the polynomial
<span class="math-container">$$P_{B,A} \equiv U^*_B - U^*_A  = \frac{I^2}{4p_1p_2} - \frac{I}{p_3} + \frac{p_1p_2}{p_3^2}.$$</span></p>
<p>Its discriminant is</p>
<p><span class="math-container">$$\Delta = \left(\frac{1}{p_3}\right)^2 - 4\frac{1}{4p_1p_2}\frac{p_1p_2}{p_3^2} = 0,$$</span></p>
<p>so the polynomial has a double root which is
<span class="math-container">$$I_{d}= \frac{1/p_3}{2/4p_1p_2} = \frac{2p_1p_2}{p_3},$$</span></p>
<p>and the polynomial can be written
<span class="math-container">$$P_{B,A} \equiv U^*_B - U^*_A  =\frac{1}{4p_1p_2}\left(I- \frac{2p_1p_2}{p_3}\right)^2 &gt; 0.$$</span></p>
<p>The inequality is strict because in the case we are considering, income <span class="math-container">$I$</span> is strictly greater than <span class="math-container">$2p_1p_2 / p_3$</span>.
So we conclude that, comparing consuming <span class="math-container">$(x_1, x_2, x_3)$</span> against just <span class="math-container">$(x_1, x_2)$</span>, we prefer the latter.</p>
<p>In all therefore, it is never optimal to consume all three goods: depending on the income level, we should either consume just <span class="math-container">$(x_1, x_2)$</span>, or just <span class="math-container">$(x_3)$</span>.</p>

CC BY-SA 4.0
2023-10-16T20:31:08.507Z
2023-10-16T20:31:08.507Z
Alecos Papadopoulos
<p>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.  </p>

<p>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.  </p>

<p>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.</p>

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

CC BY-SA 4.0
2023-10-16T00:24:37.293Z
$x_1=\frac{p_2}{p_3}$ 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 $u(4,4,8)=24<40=u(0,0,40)$.
Amit
CC BY-SA 4.0
2023-10-16T14:28:34.303Z
@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.
Alecos Papadopoulos
CC BY-SA 3.0
2015-10-04T12:03:43.627Z
2015-10-04T12:03:43.627Z
Alecos Papadopoulos
<p>I have the following utility function:</p>

<p>$$U(x_i)=x_1x_2+x_3$$</p>

<p>with budget constraint:</p>

<p>$$p_1x_1+p_2x_2+p_3x_3\leq I$$</p>

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

<p>$$x_2-\lambda p_1+M_1=0$$
$$x_1-\lambda p_2+M_2=0$$</p>

<p>$$1-\lambda p_3+M_3=0$$</p>

<p>$$p_1M_1=0$$
$$p_2M_2=0$$
$$p_3M_3=0$$
$$p_1x_1+p_2x_2+p_3x_3-I=0$$</p>

<p>$$(M_1,M_2,M_3,\lambda \geq 0)$$</p>

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

<p>$$x_1=\frac{p2}{p3}$$ and $$x_2=\frac{p_1}{p_3}$$</p>

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

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

<blockquote>
  <p>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</p>
</blockquote>

<microeconomics><consumer-theory><optimization>
Does the Marshallian demand function always include prices and income?
Übel Yildmar
# Does the Marshallian demand function always include prices and income?

(Asked by: username_0 on 2015-10-04)

I have the following utility function:

$$U(x\_i)=x\_1x\_2+x\_3$$

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

**Tags:** microeconomics, consumer-theory, optimization

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## 3 answers

### Answer by username_2 on 2015-10-04. Score: 7 | Accepted answer

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.

> **Comments:**
> **username_1**: $x_1=\frac{p_2}{p_3}$ 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 $u(4,4,8)=24<40=u(0,0,40)$.  
> **username_2**: @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.  

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### Answer by username_1 on 2023-10-15. Score: 5

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>


---

### Answer by username_2 on 2023-10-16. Score: 3

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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