# Tag Info

1 vote

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

As Michael pointed out that we cannot use Kuhn-Tucker conditions here because the objective function is not differentiable everywhere (at the boundary). Here is what we can do in light of the above: ...

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

No, you can't. The Karush-Kuhn-Tucker theorem is for functions defined on $\mathbb{R}^n$ or at least an open subset thereof and requires the function to be maximized to be differentiable on the domain....
1 vote

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

\begin{eqnarray*}\max_{x_1,x_2, x_3, x_4} & \sqrt{x_1x_2} + \sqrt{x_3x_4} \\ \text{s.t.} & \ p_1x_1 + p_2x_2+p_3x_3 +p_4x_4 \leq w \\ \text{and } & \ x_1, x_2, x_3, x_4 \geq 0\end{eqnarray*...
Accepted

### Concave utility functions solution example

This is the problem we want to solve: \begin{eqnarray*} \max_{x_1, x_2} & x_1 +\ln x_2 \\ \text{s.t.} & \ p_1 x_1 + p_2x_2 \leq w \\ \text{and} & \ x_1\geq 0, x_2>0 \end{eqnarray*} Here ...

### System of first order partial differential equation

I don't think your reasoning is correct. Some remarks: If you have an optimization problem, then you assume that you know the objective function, which in your case contains the function $u(x,y)$. As ...
Seems like the $$n_{t}(\tau+e_{t+1})=1-\frac{\tilde{c}}{z_{t}}$$ equation you don't understand is just a simple rearrangement of the first condition $$z_{t}(1-n_{t}(\tau+e_{t+1}))\geq \tilde{c}$$ if ...