<p>No, you should not use Lagrange multipliers here, but sound thinking. Suppose $x\neq y$, say for concreteness $x&lt;y$. Let $\epsilon=y-x$. Then $\min\{x,y\}=x=\min\{x,x\}=\min\{x,y-\epsilon\}.$
So the consumer could reduce her consumption of good 2, without being worse off. On the other hand for all $\delta&gt;0$, we would have $\min\{x+\delta,y-\epsilon/2\}&gt;x=\min\{x,y\}$, so the consumer could be better of by reducing the consumption of the second good and spending the freed money on the first good. In an optimum, a consumer cannot improve so optimality requires $x=y$. It is also clear that consumers improve along the $x=y$ 45° ray. So you can simply use $x=y$ as an optimality condition to be substituted into your budget constraint and bypass Lagrange multipliers.</p>

CC BY-SA 3.0
2015-01-11T10:45:33.347Z
2015-01-11T10:45:33.347Z
Michael Greinecker
<p>I am confused about a particular point regarding finding a demand function. All the problems in this practice set I am doing have involved applying the method of Lagrangian multipliers. But I am uncertain if it applies here for this problem.</p>

<p><strong>Problem Setup</strong></p>

<p>Consider a consumer with utility function $u(x,y) = \min\lbrace x,y\rbrace$. Suppose we are given wealth $w$ and prices $p_x = 1, p_y = \frac{1}{2}$. </p>

<p><strong>My Work</strong> </p>

<p>Not much to do yet. All I did was set up a budget constraint $w = xp_x +  yp_y = x + \frac{1}{2} y$.</p>

<p><strong>My Confusion</strong> </p>

<p>I was all set to setup a Lagrangian multiplier equation when suddenly I realized that my utility function is a $\min$ function. At first, I thought this function wasn't differentiable. Now, I am thinking it is not differentiable but it is partially differentiable. I am still unsure.</p>

<p><strong>My Guess</strong> </p>

<p>I suspect yes $\min$ is partially differentiable based on this thread </p>

<p><a href="https://math.stackexchange.com/questions/150960/derivative-of-the-fx-y-minx-y">https://math.stackexchange.com/questions/150960/derivative-of-the-fx-y-minx-y</a></p>

<p>But I suspect my answer will need a piecewise component or something. </p>

<p><strong>My Question</strong> </p>

<p>Are Lagrangian multipliers applicable here? If so, how do I define the Lagrangian in piecewise terms as I think I will need to do?  If it is not differentiable, how does one derive a demand function given a $\min$ or a $\max$ function? </p>

CC BY-SA 3.0
2015-11-05T03:42:37.007Z
See also this question over at Math SE: http://math.stackexchange.com/questions/605925/constrained-maximization-of-leontif-utility-function-minx-1-x-2?rq=1
EthanAlvaree
<utility>
Finding demand function given a utility min(x,y) function
Stan Shunpike
# Finding demand function given a utility min(x,y) function

(Asked by: username_0 on 2015-01-11)

I am confused about a particular point regarding finding a demand function. All the problems in this practice set I am doing have involved applying the method of Lagrangian multipliers. But I am uncertain if it applies here for this problem.

**Problem Setup**

Consider a consumer with utility function $u(x,y) = \min\lbrace x,y\rbrace$. Suppose we are given wealth $w$ and prices $p\_x = 1, p\_y = \frac{1}{2}$.

**My Work**

Not much to do yet. All I did was set up a budget constraint $w = xp\_x + yp\_y = x + \frac{1}{2} y$.

**My Confusion**

I was all set to setup a Lagrangian multiplier equation when suddenly I realized that my utility function is a $\min$ function. At first, I thought this function wasn't differentiable. Now, I am thinking it is not differentiable but it is partially differentiable. I am still unsure.

**My Guess**

I suspect yes $\min$ is partially differentiable based on this thread

<https://math.stackexchange.com/questions/150960/derivative-of-the-fx-y-minx-y>

But I suspect my answer will need a piecewise component or something.

**My Question**

Are Lagrangian multipliers applicable here? If so, how do I define the Lagrangian in piecewise terms as I think I will need to do? If it is not differentiable, how does one derive a demand function given a $\min$ or a $\max$ function?

**Tags:** utility
> **Comments:**
> **username_1**: See also this question over at Math SE: http://math.stackexchange.com/questions/605925/constrained-maximization-of-leontif-utility-function-minx-1-x-2?rq=1  

---

## 1 answers

### Answer by username_2 on 2015-01-11. Score: 11 | Accepted answer

No, you should not use Lagrange multipliers here, but sound thinking. Suppose $x\neq y$, say for concreteness $x0$, we would have $\min\{x+\delta,y-\epsilon/2\}>x=\min\{x,y\}$, so the consumer could be better of by reducing the consumption of the second good and spending the freed money on the first good. In an optimum, a consumer cannot improve so optimality requires $x=y$. It is also clear that consumers improve along the $x=y$ 45° ray. So you can simply use $x=y$ as an optimality condition to be substituted into your budget constraint and bypass Lagrange multipliers.


---

