# How to find marshallian demand for Leontief Utility with 3 goods. u(x 1 ​ ,x 2 ​ ,x 3 ​)=min{2x1 + x3, x2/2}

I have a utility function

$$u(x_1, x_2, x_3) = \min \{2x_1 + x_3, x_2/2\}$$

I would have assumed that the relationship established is $$2x_1 + x_3 = x_2/2$$ but my solution manual has it as $$x_1 + 2x_2 = x_3/2$$

I'm very confused

• You are right, the optimal choice i.e. the Marshallian demand in this problem satisfy $2x_1+x_3=\frac{x_2}{2}$.
– Amit
Commented Aug 20 at 17:59
• Your solution manual appears to be wrong, unless there's a typo or the suggested solution is a simplification after factoring in the prices (which have not been provided here). Commented Aug 20 at 18:04
• Yeah I was worried about that. My workout reaches as far as rearranching the relationship and substituting x2 = 4x1 + 2x3 into the endowment function/ ("budget constraint") to obtain m = p1x2 + 4p2x1 + 2p2x3 + p3x3 but I'm not sure how to progress from that point Commented Aug 20 at 19:50

$$\begin{eqnarray*}\max_{(x_1,x_2,x_3)\in\mathbb{R}^3_+} & \min\left(2x_1+x_3,\frac{x_2}{2}\right) \\ \text{s.t. } & p_1x_1+p_2x_2+p_3x_3\leq M\end{eqnarray*}$$ where $$p_1>0$$, $$p_2>0$$, $$p_3>0$$, and $$M\geq 0$$ are given.
Solving this problem, we get the Marshallian demand as $$\begin{eqnarray*}(x_1^d,x_2^d,x_3^d)(p_1,p_2,p_3,M)\in \begin{cases} \left\{\left(\dfrac{M}{p_1+4p_2},\dfrac{4M}{p_1+4p_2},0\right)\right\} & \text{if } p_1<2p_3 \\ \left\{\left(0,\dfrac{2M}{2p_2+p_3},\dfrac{M}{2p_2+p_3}\right)\right\} & \text{if } p_1>2p_3 \\ \left\{\left(\dfrac{\theta M}{p_1+4p_2},\dfrac{4M}{p_1+4p_2},\dfrac{(1-\theta)M}{2p_2+p_3}\right)|\theta\in[0,1]\right\} & \text{if } p_1=2p_3\end{cases}\end{eqnarray*}$$
• Because we need to choose between $x_1$ and $x_3$ since the two are substitutes, and here the expression is $2x_1+x_3$.