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Consider

$\max_{x_1, x_2, x_3, x_4} u(x) = \sqrt{x_1 x_2} + \sqrt{x_3 x_4}$

s.t. $\; p_1x_1 + p_2x_2 + p_3x_3 + p_4x_4 \le w$

I know we can solve the max problem through separately considering case(i): $x_1, x_2 > 0$ and $x_3 = x_4 = 0$; and case (ii) $x_1 = x_2 = 0$ and $x_3, x_4 > 0$.

But is it possible to solve the whole optimization problem through the Kuhn–Tucker method?

We can write down the Lagrangian $L(x,\lambda)=\sqrt{x_1x_2}+\sqrt{x_3x_4}+\lambda(w-p_1x_1-p_2x_2-p_3x_3-p_4x_4)$

with the complementary slackness conditions:

$\frac{\partial L}{\partial x_1}=\frac{1}{2}\sqrt{\frac{x_2}{x_1}}-\lambda p_1 \le 0,\quad x_1 \ge 0, \quad \text{and}\quad x_1 \frac{\partial L}{\partial x_1}=0$.

$\frac{\partial L}{\partial x_2}=\frac{1}{2}\sqrt{\frac{x_1}{x_2}}-\lambda p_2 \le 0,\quad x_2 \ge 0, \quad \text{and}\quad x_2 \frac{\partial L}{\partial x_2}=0$.

$\frac{\partial L}{\partial x_3}=\frac{1}{2}\sqrt{\frac{x_4}{x_3}}-\lambda p_3 \le 0,\quad x_3 \ge 0, \quad \text{and}\quad x_3 \frac{\partial L}{\partial x_3}=0$.

$\frac{\partial L}{\partial x_4}=\frac{1}{2}\sqrt{\frac{x_3}{x_4}}-\lambda p_4 \le 0,\quad x_4 \ge 0, \quad \text{and}\quad x_4 \frac{\partial L}{\partial x_4}=0$.

However, when guessing say, $x_1 = 0$ and $x_2,x_3,x_4>0$, $\lim_{x_1 \to 0}\frac{\partial L}{\partial x_1} \to \infty$, which does not satisfy the complementary slackness $x_1 \frac{\partial L}{\partial x_1} = 0$.

I do not know whether we can use the Kuhn–Tucker method to solve this optimization problem? And if not, what are the reasons?

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    $\begingroup$ Here's a good trick for any microeconomic optimization problem: if the agent's preferences are locally non-satiated, then Walras' Law will hold. That is, $p x = w$. Also note that monotonic preferences are locally non-satiated, and as your objective function is weakly increasing in each of its arguments, you have weakly monotonic preferences. $\endgroup$ – NBm424 Oct 30 '18 at 1:59
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Why do you want to guess that some $x$ is zero? What is the problem of having all four goods strictly positive at the solution?

In fact, due to the presence of the square root in the first order conditions, the solution looks straightforward, since the inequality cannot hold (and assuming exhaustion of the budget).


ADDENDUM
Given the f.o.c's provided by the OP, all $x$'s have to be strictly positive at the solution, since otherwise we have division by zero and/or the undetermined form $0/0$.

This in turn implies that $\partial L/\partial x_i = 0$ at the solution, which now becomes easy to compute.

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  • $\begingroup$ It's because the solutions are (i) if $p_1p_2 < p_3 p_4$, then $x_1 = \frac{M}{2p_1}$, $x_2 = \frac{M}{2p_2}$, $x_3 =x_4 =0$; (ii) if $p_1p_2 > p_3 p_4$, then $x_3 = \frac{M}{2p_3}$, $x_4 = \frac{M}{2p_4}$, $x_1 =x_2 =0$; and (iii) with an interior solution, when $p_1p_2 = p_3 p_4$. $\endgroup$ – Yun Oct 30 '18 at 5:39

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