Consider the max problem
\begin{align*} & max_{w_s,w_f} 0.75(7-w_s)+0.25(2-w_f) \\ & \text{subject to} \\ & 0.75 \sqrt{w_s} + 0.25 \sqrt{w_f} - 1 \geq \bar{U} \\ & 0.25 \left( \sqrt{w_s} - \sqrt{w_f} \right) \geq 1 \end{align*}
How can I then solve the problem when I have two by using Lagrange? I am mostly used to only having one Lagrange multipliers, thus having two constraints and therefore two multipliers confuses me. This is what I have done so far:
$$\mathcal{L} = 0.75(7-w_s)+0.25(2-w_f)+\lambda(0.25(\sqrt{w_s}-\sqrt{w_f})-1)+\mu(0.75 \cdot \sqrt{w_s} + 0.25 \cdot \sqrt{w_f} -1 - \bar{U}$$
$$\frac{\partial}{\partial w_s} = -0.75 +\frac{\lambda \cdot 0.25}{2 \cdot \sqrt{w_s}} + \frac{\mu \cdot 0.75}{2 \cdot \sqrt{w_s}} = 0$$ $$\frac{\partial}{\partial w_f} = -0.25 -\frac{\lambda \cdot 0.25}{2 \cdot \sqrt{w_f}} + \frac{\mu \cdot 0.25}{2 \cdot \sqrt{w_f}} = 0$$ $$\lambda(0.25( \sqrt{w_s} - \sqrt{w_f}) -1 ) = 0 $$ $$\mu(0.75 \cdot \sqrt{w_s} + 0.25 \cdot \sqrt{w_f} - 1 - \bar{U}) = 0$$
TIA.
To solve these problems, you can use the fact that either the multiplier is zero or the constraint is binding (or both). Assuming that the constraints are binding leaves you with the following system of equations $$ \begin{split} -0.75 +\frac{\lambda \cdot 0.25}{2 \cdot \sqrt{w_s}} + \frac{\mu \cdot 0.75}{2 \cdot \sqrt{w_s}} &= 0 \\ -0.25 -\frac{\lambda \cdot 0.25}{2 \cdot \sqrt{w_f}} + \frac{\mu \cdot 0.25}{2 \cdot \sqrt{w_f}} &= 0 \\ 0.25( \sqrt{w_s} - \sqrt{w_f}) -1 &= 0 \\ 0.75 \cdot \sqrt{w_s} + 0.25 \cdot \sqrt{w_f} - 1 - \bar{U} &= 0. \end{split} $$
The last two can be solved for $w_s,w_f$ (solution is $w_s=(\bar U +1)^2$, $w_f=(\bar U -3)^2$). To check whether the assumption of binding constraints is justified, you can solve for the multipliers, they need to be non negative.