Given $U : \mathbb{R^2} \to \mathbb{R}$ is monotone and quasi-concave, consider the following problem :
$$\max_{(x,y) \ \in \ \mathbb{R}^2}[U(x,y)] \text{ subject to } p_1 x + p_2 y \leq M ; \ (p_1, p_2 > 0)$$
1. If a level curve of $U$ is tangent to $p_1 x + p_2 y = M$ at some point $(a,b)$ on such that the slope of the tangent is $-\frac{p_1}{p_2}$, then show that it is an optimal point.
I have seen the tangency condition to be written as $\displaystyle \frac{df(x)}{dx} = -\frac{\frac{\partial U(x,y)}{\partial x}}{\frac{\partial U(x,y)}{\partial y}} = -\frac{p_1}{p_2}$ where $f: (a-\epsilon, a+\epsilon) \to \mathbb{R}$ describes the level curve around the point $(a,b)$.
2. Is it possible to have $\frac{df(x)}{dx} = -\frac{p_1}{p_2}$ but one or both of $\frac{\partial U(x,y)}{\partial x}$ and $\frac{\partial U(x,y)}{\partial y}$ are not defined?
My attempt:
If $U$ is differentiable at all points, then the Karush-Kuhn-Tucker condition is satisfied and (1) follows. For (2), each term is defined due to our assumption.
The problem lies when $U$ is not differentiable everywhere. In that case, how do we show that the point of tangency is an optimal point? I don't know if the tangency condition requires all the derivatives in (2) to be defined in the interval around $(a,b)$; assume differentiability of $U$ in that particular region around $(a,b)$ only if that is a necessity.
Definitions:
Quasi-concavity of $U$ :
(Definition $1$) If $U(r) \geq U(s)$ for $r,s \in \mathbb{R}^2$, then $U(\lambda r + (1-\lambda)s) \geq U(s) \ \forall \lambda \in [0,1]$.
(Definition $2$) The set $S_\alpha := \{ (x,y) : U(x,y) \geq \alpha\}$ is convex for all values of $\alpha$.
Monotonicity of $U$ :
- $r \geq s \implies U(r) \geq U(s)$ where $r \geq s$ means $(r_1 \geq r_2) \land (s_1 \geq s_2)$ for $r = (r_1, r_2)$ and $s = (s_1, s_2)$.