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.

• $ax^2+by^2$ is not concave, unless $a,b<0$. Sep 27 '18 at 15:50

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-

$$\begin{gather*} x_1= \begin{cases} \frac{w}{p_1} - 1,& \text{if } w>p_1\\ 0, & \text{otherwise} \end{cases}\\ x_2= \begin{cases} \frac{p_1}{p_2} ,& \text{if } w>p_1\\ \frac{w}{p_2}, & \text{otherwise} \end{cases} \end{gather*}$$

The corner solutions in this case are obtained whenever $$w\leq p_1$$.