Use Lagrange method to solve the constrained OP $\max_{x_{1}, x_{2}} 2x_{1} + 3 x_{2}$

Question

subject to $\sqrt{x_{1}} + \sqrt{x_{2}} = 5$.

I set up the Lagrangian first $\wedge (x_{1},x_{2}, \lambda) = 2 x_{1} + 3 x_{2} + \lambda\left ( 5 - \sqrt{x_{1}} - \sqrt{x_{2}} \right )$.

Then obtain the following first-order conditions:

$$\frac{\partial \wedge }{\partial x_{1}} = 2 - \frac{\lambda}{2\sqrt{x_{1}}} = 0$$

$$\frac{\partial \wedge }{\partial x_{2}} = 3 - \frac{\lambda}{2\sqrt{x_{2}}} = 0$$

$$\frac{\partial \wedge }{\partial \lambda} = 5 - \sqrt{x_{1}} - \sqrt{x_{2}} = 0$$

Then I solve the above equations and find that $x_{1}=9$, $x_{2} = 4$, and $\lambda = 12$.

However the answer says that for the optimal solution $x_{1}=0$ and $x_{2}=25$.

I am not sure how they find this. Also, how do I check the constraint qualification the proper way?

For CQ, how do I show that the constraint is full rank?

I have $$ J = \begin{bmatrix} \frac{1}{2 \sqrt{x_{1}}} & \frac{1}{2 \sqrt{x_{2}}} \end{bmatrix}$$

Then I am not sure how to infer that the Jacobian matrix is full rank.

Answer 1

If $x_1=0$ and $x_2=25$ we get a value $75$.

We'll prove that it's a maximal value.

Indeed, we need to prove that $$2x_1+3x_2\leq75$$ or $$2x_1+3x_2\leq3(\sqrt{x_1}+\sqrt{x_2})^2$$ or $$2x_1+3x_2\leq3x_1+6\sqrt{x_1x_2}+3x_2,$$ which is obvious.

Done!

Answer 2

one can do the following from $$\sqrt{x_2}=5-\sqrt{x_1}$$ we get by squaring $$x_2=25-10\sqrt{x_1}+x_1$$ and the objective function is now $$f(x_1)=2x_1+3(25-10\sqrt{x_1}+x_1)$$ simplifying $$f(x_1)=5x_1-30\sqrt{x_1}+75$$ $$f'(x_1)=5-15x_1^{-1/2}$$ and $$f''(x_1)=\frac{15}{2}x_1^{-3/2}>0$$ gives a minimum