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

Question

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

Answer 1

Given the utility maximisation problem,

\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*}