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How can I derive Hicksian demand, when from the FOC I only get $\frac{p_x}{p_y} = \frac13$ without the usual x & 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

$$p_x \cdot x + p_y \cdot y \qquad\text{s.t.}\qquad x + 3y ≥ U$$

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  • $\begingroup$ Is the constraint $x + 3y \geq U$ or $x + 3y > U$? $\endgroup$ – Amit Aug 24 at 21:07
  • $\begingroup$ it's ≥ sign and I corrected the text, thanks $\endgroup$ – Svit Valenčič Aug 25 at 8:03
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Expenditure minimization problem in the question is as follows : \begin{eqnarray*} \min_{x\geq 0, y\geq 0} & \ \ p_Xx + p_Yy \\ \text{s.t.}& \ \ x + 3y \geq U \end{eqnarray*} where $p_X > 0$, $p_Y > 0$ 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} & \ \ p_Xx + p_Yy \\ \text{s.t.}& \ \ 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} & \ \ 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 & \text{if } p_X < \frac{p_Y}{3} \\ \in [0, U] & \text{if } p_X = \frac{p_Y}{3} \\ = 0 & \text{if } p_X > \frac{p_Y}{3}\end{cases} \end{eqnarray*}

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enter image description here

To start with, plot the line for which \begin{equation} x + 3y = U \end{equation} The red line in the graph above is this line. For simplicity, I've assumed U = 9 in this case but the solution would work for any U > 0.

The region to the right of this line would satisfy your constraint but since we are interested in minimising our cost, we would want our budget line to intersect this particular line.

Case 1: The budget line is flatter (black line) as compared to the constraint line, i.e. Px/Py < 1/3. To minimise cost, the line should be such that it intersects the constraint line at the right corner where x = U.

Case 2: The budget line is steeper (purple line) as compared to the constraint line, i.e, Px/Py > 1/3. To minimise cost, the line should be such that it intersects the constraint line at the left corner where x = 0.

Case 3: If the slope of the budget line is equal to the slope of the constraint line, i.e. if Px/Py = 1/3, any point on the constraint line would be a solution point. x can vary anywhere from 0 to U.

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  • $\begingroup$ @svit notice that these three cases in this answers correspond directly to the other answers’ solution(s) $\endgroup$ – Brennan Aug 26 at 14:52

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