$max - \frac{1}{x} - \frac{1}{y}$ s.t. $2x + y ≤ 10, x ≥ 0, y ≥ 0$

Question

$\max - \frac{1}{x} - \frac{1}{y}$

s.t. $2x + y ≤ 10,\quad x ≥ 0,\quad y ≥ 0$

I set up the lagrangian and take FOC.

$ \frac{\frac{1}{x^2}}{\frac{1}{y^2}}$=$\frac{2}{1}$

$y=\sqrt2$ x Substitute in λ partial derivative and we get $x* = 2.92$ and $y* = 4.14$?

I get $y^* =4.14$ and $x^*=2.92$. Am I correct? Why don't I need to add lagrangrian for the other two constraints as below?

$L= - \frac{1}{x} - \frac{1}{y}$ + $λ_1(10-2x-y)+λ_2x+λ_3y$?

Answer 1

You do not need to add the positivity constraints because here those constraints are unbinding i.e. the candidates for your optimum values are positive $\Rightarrow$ your constraint is unbinding $\Rightarrow$ $\lambda_{2}$ & $\lambda_{3}$ are zero. Therefore, they are not added.

In general case when we use Lagrange Multiplier method and derive Kuhn Tucker Conditions then we test our candidates for optimum values on 2 conditions related to the positivity constraint $x\leq0$ and $\lambda_{2}x=0$. For further information on this topic, I would suggest you to read Chapter 18 of Simon & Blume "Mathematics for Economists" & Chapter on Optimization from "Intermediate Micro" by Thomas Nechyba.