<p>If <span class="math-container">$\rho &lt; 1$</span> then the marginal utility (say with respect to <span class="math-container">$x_1$</span> equals):
<span class="math-container">$$
\frac{\partial U}{\partial x_1}(x_1, x_2) = x_1^{\rho - 1} (x_1^\rho + x_2^\rho)^{\frac{1}{\rho}- 1},$$</span>
This tends to <span class="math-container">$+\infty$</span> as <span class="math-container">$x_1$</span> tends to zero.</p>
<p>As such, it will never be optimal to set <span class="math-container">$x_1$</span> equal to zero.</p>

CC BY-SA 4.0
2024-02-20T14:02:13.530Z
I do not understand this explanation fully. I get that big marginal utility incentivizes me to purchase another product stronger than small marginal utility. To be clear, on new hypothetical example, if marginal utility increased to $+ \infty$ as $x_1$ went, for instance, to $-2$, this situation would allow for setting $x_1 = 0$?
Athaeneus
CC BY-SA 4.0
2024-02-20T15:01:27.770Z
The intuition is that setting $x_1 = 0$ is never optimal because increasing $x_1$ a tiny bit (with a corresponding marginal decrease in $x_2$) would lead to an infinite increase in utility. In principle, you can always use the KT conditions and then verify that the constraints $x_1 \ge 0$ and $x_2 \ge 0$ are never binding.
tdm
CC BY-SA 4.0
2024-02-20T12:59:50.730Z
2024-02-20T12:59:50.730Z
tdm
<p>Let's have the following problem:</p>
<p><span class="math-container">$$U(\boldsymbol{x}) = \left(x_1^\rho + x_2^\rho \right)^{\frac{1}{\rho}} \qquad s.t. \qquad P_1 x_1 + P_2 x_2 \leq M$$</span></p>
<p>Is it optimal to solve this problem via pure Lagrangian:</p>
<p><span class="math-container">$$\mathscr{L} = \left(x_1^\rho + x_2^\rho \right)^{\frac{1}{\rho}} + \lambda \left( M - P_1 x_1 - P_2 x_2 \right)  $$</span></p>
<p>or Lagrangian with non-negativity constraints (or KKT):</p>
<p><span class="math-container">$$\mathscr{L} = \left(x_1^\rho + x_2^\rho \right)^{\frac{1}{\rho}} + \lambda \left( M - P_1 x_1 - P_2 x_2 \right) + \mu_1 x_1 +  \mu_2 x_2$$</span></p>
<p>The thing is that analyzing perfect substitutes requires you to use non-negativity constraints to obtain the correct solution.</p>
<p>However, for instance, what if <span class="math-container">$\rho = 0.9$</span> instead of <span class="math-container">$\rho = 1$</span>? Can we say that the optimal solution would always be interior such that we do not need to consider the non-negativity constraints?</p>

CC BY-SA 4.0
2024-02-20T16:32:36.650Z
In principle, to mathematical rigor, the condition $\geq0 $ should be set from the beginning, because  powers with rational exponent in $\mathbb{R}$ are not defined for negative numbers (only If we have an integer odd number the root is defined also for negative numbers). Otherwise, we can go to complex numbers and complex functions. But, as tdm says, this non negativity constraints are redundant in your CES case, except for $\rho=1$
BakerStreet
<microeconomics><mathematical-economics><lagrangian>
CES function: Lagrangian or Kuhn-Tucker
Athaeneus
# CES function: Lagrangian or Kuhn-Tucker

(Asked by: username_0 on 2024-02-20)

Let's have the following problem:

$$U(\boldsymbol{x}) = \left(x\_1^\rho + x\_2^\rho \right)^{\frac{1}{\rho}} \qquad s.t. \qquad P\_1 x\_1 + P\_2 x\_2 \leq M$$

Is it optimal to solve this problem via pure Lagrangian:

$$\mathscr{L} = \left(x\_1^\rho + x\_2^\rho \right)^{\frac{1}{\rho}} + \lambda \left( M - P\_1 x\_1 - P\_2 x\_2 \right) $$

or Lagrangian with non-negativity constraints (or KKT):

$$\mathscr{L} = \left(x\_1^\rho + x\_2^\rho \right)^{\frac{1}{\rho}} + \lambda \left( M - P\_1 x\_1 - P\_2 x\_2 \right) + \mu\_1 x\_1 + \mu\_2 x\_2$$

The thing is that analyzing perfect substitutes requires you to use non-negativity constraints to obtain the correct solution.

However, for instance, what if $\rho = 0.9$ instead of $\rho = 1$? Can we say that the optimal solution would always be interior such that we do not need to consider the non-negativity constraints?

**Tags:** microeconomics, mathematical-economics, lagrangian

> **Comments:**
> **username_1**: In principle, to mathematical rigor, the condition $\geq0 $ should be set from the beginning, because  powers with rational exponent in $\mathbb{R}$ are not defined for negative numbers (only If we have an integer odd number the root is defined also for negative numbers). Otherwise, we can go to complex numbers and complex functions. But, as username_2 says, this non negativity constraints are redundant in your CES case, except for $\rho=1$  

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## 1 answers

### Answer by username_2 on 2024-02-20. Score: 2 | Accepted answer

If $\rho < 1$ then the marginal utility (say with respect to $x\_1$ equals):
$$
\frac{\partial U}{\partial x\_1}(x\_1, x\_2) = x\_1^{\rho - 1} (x\_1^\rho + x\_2^\rho)^{\frac{1}{\rho}- 1},$$
This tends to $+\infty$ as $x\_1$ tends to zero.

As such, it will never be optimal to set $x\_1$ equal to zero.

> **Comments:**
> **username_0**: I do not understand this explanation fully. I get that big marginal utility incentivizes me to purchase another product stronger than small marginal utility. To be clear, on new hypothetical example, if marginal utility increased to $+ \infty$ as $x_1$ went, for instance, to $-2$, this situation would allow for setting $x_1 = 0$?  
> **username_2**: The intuition is that setting $x_1 = 0$ is never optimal because increasing $x_1$ a tiny bit (with a corresponding marginal decrease in $x_2$) would lead to an infinite increase in utility. In principle, you can always use the KT conditions and then verify that the constraints $x_1 \ge 0$ and $x_2 \ge 0$ are never binding.  

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