How to derive Hicksian demand? How can I derive Hicksian demand, when from the FOC I only get $\frac{p_x}{p_y} = \frac13$ without the usual x &amp; y. So they cannot be derived directly from FOC, but if I plug the price relation into the budget constraint $I =p_x \cdot x + p_y \cdot y$ I get the income in the demand function, so this is Marshallian demand. Plugging the relation in expenditure function, obtained from the indirect utility function also doesn't lead to the Hicksian demand (that I obtained via Shephard's lemma and equals $h_x = U + x + 3y$).

The problem is to minimize

$$p_x \cdot x + p_y \cdot y \qquad\text{s.t.}\qquad x + 3y ≥ U$$
Expenditure minimization problem in the question is as follows :
\begin{eqnarray*} \min_{x\geq 0, y\geq 0} &amp; \ \ p_Xx + p_Yy \\ \text{s.t.}&amp; \ \ x + 3y \geq U \end{eqnarray*}
where $p_X &gt; 0$, $p_Y &gt; 0$ and $U \geq 0$ are given.
Since prices are positive, the cost minimizing choice will satisfy the condition that $x + 3y = U$. So, we can rewrite the above problem as :
\begin{eqnarray*} \min_{x\geq 0, y\geq 0} &amp; \ \ p_Xx + p_Yy \\ \text{s.t.}&amp; \ \ x + 3y = U \end{eqnarray*}
We can now substitute the value of $y = \frac{U-x}{3}$ in the objective and rewrite the problem as :
\begin{eqnarray*} \min_{0 \leq x \leq U} &amp; \ \ p_Xx + p_Y\left(\dfrac{U-x}{3}\right)  \end{eqnarray*}

Since the objective ${\color{red} {p_Xx + p_Y\left(\frac{U-x}{3}\right) =\frac{p_YU}{3} + \left(p_X- \frac{p_Y}{3}\right)x }}$  is linear in $x$, solution is straight forward :
\begin{eqnarray*} x^h(p_X, p_Y, U)  \begin{cases} = U &amp; \text{if } p_X &lt; \frac{p_Y}{3} \\ \in [0, U] &amp; \text{if } p_X = \frac{p_Y}{3} \\ = 0 &amp; \text{if } p_X &gt; \frac{p_Y}{3}\end{cases} \end{eqnarray*}