<p><a href="https://i.sstatic.net/HgRQn.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HgRQn.png" alt="enter image description here"></a></p>

<p>To start with, plot the line for which <span class="math-container">\begin{equation}
x + 3y = U
\end{equation}</span>
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. </p>

<p>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.  </p>

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

<p><em>Case 2:</em> 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.</p>

<p><em>Case 3:</em> 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. </p>

CC BY-SA 4.0
2019-08-26T14:52:13.900Z
@svit notice that these three cases in this answers correspond directly to the other answers’ solution(s)
Brennan
CC BY-SA 4.0
2019-08-26T12:02:24.980Z
2019-08-26T12:02:24.980Z
Rayhan Ali
<p>Expenditure minimization problem in the question is as follows :
<span class="math-container">\begin{eqnarray*} \min_{x\geq 0, y\geq 0} &amp; \ \ p_Xx + p_Yy \\ \text{s.t.}&amp; \ \ x + 3y \geq U \end{eqnarray*}</span>
where <span class="math-container">$p_X &gt; 0$</span>, <span class="math-container">$p_Y &gt; 0$</span> and <span class="math-container">$U \geq 0$</span> are given.
Since prices are positive, the cost minimizing choice will satisfy the condition that <span class="math-container">$x + 3y = U$</span>. So, we can rewrite the above problem as :
<span class="math-container">\begin{eqnarray*} \min_{x\geq 0, y\geq 0} &amp; \ \ p_Xx + p_Yy \\ \text{s.t.}&amp; \ \ x + 3y = U \end{eqnarray*}</span>
We can now substitute the value of <span class="math-container">$y = \frac{U-x}{3}$</span> in the objective and rewrite the problem as :
<span class="math-container">\begin{eqnarray*} \min_{0 \leq x \leq U} &amp; \ \ p_Xx + p_Y\left(\dfrac{U-x}{3}\right)  \end{eqnarray*}</span></p>

<p>Since the objective <span class="math-container">${\color{red} {p_Xx + p_Y\left(\frac{U-x}{3}\right) =\frac{p_YU}{3} + \left(p_X- \frac{p_Y}{3}\right)x }}$</span>  is linear in <span class="math-container">$x$</span>, solution is straight forward :
<span class="math-container">\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*}</span></p>

CC BY-SA 4.0
2019-08-25T10:48:02.157Z
2019-08-25T10:48:02.157Z
Amit
<p>How can I derive Hicksian demand, when from the FOC I only get <span class="math-container">$\frac{p_x}{p_y} = \frac13$</span> 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 <span class="math-container">$I =p_x \cdot x + p_y \cdot y$</span> 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 <span class="math-container">$h_x = U + x + 3y$</span>).</p>

<p>The problem is to minimize</p>

<p><span class="math-container">$$p_x \cdot x + p_y \cdot y \qquad\text{s.t.}\qquad x + 3y ≥ U$$</span></p>

CC BY-SA 4.0
2019-08-24T21:07:38.337Z
Is the constraint $x + 3y \geq U$ or $x + 3y > U$?
Amit
CC BY-SA 4.0
2019-08-25T08:03:48.553Z
it's ≥ sign and I corrected the text, thanks
Svit
<microeconomics>
How to derive Hicksian demand?
Svit
# How to derive Hicksian demand?

(Asked by: username_0 on 2019-08-24)

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

$$p\_x \cdot x + p\_y \cdot y \qquad\text{s.t.}\qquad x + 3y ≥ U$$

**Tags:** microeconomics
> **Comments:**
> **username_1**: Is the constraint $x + 3y \geq U$ or $x + 3y > U$?  
> **username_0**: it's ≥ sign and I corrected the text, thanks  

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## 2 answers

### Answer by username_1 on 2019-08-25. Score: 4 | Accepted answer

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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### Answer by username_2 on 2019-08-26. Score: 2

[![enter image description here](https://i.sstatic.net/HgRQn.png)](https://i.sstatic.net/HgRQn.png)

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.

> **Comments:**
> **username_3**: @svit notice that these three cases in this answers correspond directly to the other answers’ solution(s)  

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