Concave utility functions corner solution explanation I seem to not be getting this. Could someone explain me the mathematical way to show a concave utility function [like (ax^2+by^2)] subject to a budget constraint has a corner solution. I get the economic intuition behind it but not the mathematical part behind it.
Intuitively, one can see that the corner solutions lie on the highest possible indifference curve(for concave preferences). Thus, utility obtained will be highest on those points. The following figure(courtesy-Google Images) might help one to visualize the above statement.

Mathematically, it is a bit tough to visualize whats going on. However, the Kuhn Tucker conditions can tell you whether a corner solution will exist or not(note that concavity/convexity of preferences are of no use while using Kuhn-Tucker Theorem, however one needs to check for differentiability of the objective function/constraints). First of all, according to Weierstrass' Theorem, it is guaranteed that an optimum point for $U(x_1,x_2)$ exists in the budget set $B(p_1,p_2,w)=\{(x_1,x_2):p_1 x_1 + p_2 x_2 \leq w\}$ (only if $U(x_1,x_2)$ is continuous in the given domain,i.e; the budget set). This tells us that we cannot rule out any solution/optimal point where either $x_1=0$ or $x_2=0$, or both. Now, to check whether a corner solution exists or not, we can proceed as follows. Consider the maximization problem
\begin{gather*}
\max\ U(x_1,x_2); \\ s.t.\ p_1x_1+p_2x_2\leq w;\ x_1\geq 0, x_2\geq 0.\\
\end{gather*}
The Kuhn Tucker Lagrangian is given as
\begin{align}
L = U(x_1,x_2)+ \lambda(w-p_1x_1+p_2x_2)+ \mu_1 x_1+ \mu_2 x_2.
\end{align}
The Kuhn Tucker conditions are given as-
\begin{gather*}
\frac{\partial L}{\partial x_1}\leq0\tag{1}\\ \frac{\partial L}{\partial x_2}\leq0\tag{2}\\ \frac{\partial L}{\partial \lambda}\leq0 ,\ \frac{\partial L}{\partial \mu_1}\leq0\ ,\frac{\partial L}{\partial \mu_2}\leq0\tag{3}\\ \lambda\frac{\partial L}{\partial \lambda}=0\tag{4}\\ \mu_1\frac{\partial L}{\partial \mu_1}=0\ ,\mu_2\frac{\partial L}{\partial \mu_2}=0\tag{5}
\end{gather*}
The inequality signs in conditions $(1)$ and $(2)$ can be replaced with the equality sign, if the constraints are binding(you can read more about binding constraints/constraint qualifications from Simon and Blume).
However, the given problem of corner solutions can be understood from the conditions in $(5)$, which say that $\mu_1 x_1=0$ and $\mu_2 x_2=0$. These equations tell us that either any one(or both) of the $\mu_i=0$ or any one(or both) of the $x_i = 0$. By considering all the above cases one by one, we can figure out whether a corner solution will exist or not.
For example, if the utility function is given as $U(x_1,x_2) = x_1 + \ln(x_2)$, the solutions obtained by using Kuhn Tucker theorem are-
\begin{gather*}
    x_1= 
\begin{cases}
    \frac{w}{p_1} - 1,&amp; \text{if } w&gt;p_1\\
    0,              &amp; \text{otherwise}
\end{cases}\\
x_2= 
\begin{cases}
    \frac{p_1}{p_2} ,&amp; \text{if } w&gt;p_1\\
    \frac{w}{p_2},              &amp; \text{otherwise}
\end{cases}
\end{gather*}
The corner solutions in this case are obtained whenever $w\leq p_1$.