# Can $u(x) = \sqrt{x_1 x_2} + \sqrt{x_3 x_4}$ be solved by Kuhn–Tucker conditions?

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?

• 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. – NBm424 Oct 30 '18 at 1:59

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?
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.
• 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 $p_1p_2 = p_3 p_4$. – Yun Oct 30 '18 at 5:39