# Any interior solution for $u(x,y) = min\left \{ x,y \right \}^{2} + max\left \{ x,y \right \}$?

Will all the solutions be in the corner or will the cusp in the middle give us any interior solution? This is by the intersection of the budget line.

I am getting this type of a shape: But I am not sure if there will be interior solution. If there is we can't use differentiation because of the cusp. I think that there will only be corner solutions but it depends on the shape of the indifference curve.

Because the two segments of the IC bend away from the origin, the only possible solutions are at the corners or at $$x=y$$.

Write $$p_x$$ and $$p_y$$ for the prices, and $$M$$ for the budget.

Suppose we try to set up a corner solution in which only one of the two goods is consumed. Since utility is symmetric, it makes sense to spend all of your money on whichever good is cheaper. Suppose this is good $$x$$. We then have $$x=\frac{M}{p_x}>0=y$$ and $$u=x$$.

The best $$x=y$$ bundle that can be afforded is such that $$x p_x+x p_y=M\iff x=y=\frac{M}{p_x+p_y}$$. Utility is then $$\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y}.$$

You should easily be able to compute a condition on $$M$$ such that the bundle with $$x=y$$ is optimal: $$M>\frac{p_y(p_x+p_y)}{p_x}.$$

You can see this dependence on M if we look at a contour plot of the indifference curves: • Thanks for the nice explanation with the graph included. Btw, so is the case of $x=y$ an interior solution? From the graph it appears as though you can have interior solution at $x=y$? – OGC Nov 27 '18 at 10:30
• @OGC It depends what you mean by 'interior solution' (the definition of that term is a bit fuzzy). The solution is interior in the sense that both goods are consumed in positive quantities (so long as $M$ is large enough), but non-interior in the sense that the Marginal Rate of Substitution does not equal the relative price as in the typical textbook solution to utility maximization. – Ubiquitous Nov 27 '18 at 10:33
• @Ubiquitious I see. By interior I meant $x$ and $y$ being both positive. However, I do not quite follow how you were able to obtain the price condition for $x=y$ case. Normally, I would try to find such a condition from the indirect utility function, $v\left ( p,u \right ) = max \left \{ \frac{M}{p_{x}}, \frac{M}{p_{y}},\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y} \right \}$. – OGC Nov 27 '18 at 11:28
• @OGC The budget constraint is $xp_x+yp_y=M$. I am looking for the best bundle with $x=y$, so I make this substitution to yield $xp_x+xp_y=M$. Then I simply solve to find the quantity $x$ that expends the full income (the quantity of $y$ in the bundle is the same by definition). – Ubiquitous Nov 27 '18 at 12:35
• @OGC Or did you mean the last condition? That is found by setting $\max_x u(x,0)<\max_x u(x,x)$ and solving for $M$: $$\frac{M}{p_X}<\frac{M^2}{(p_x+p_y)^2}+\frac{M}{p_x+p_y}\implies M>\frac{p_y(p_x+p_y)}{p_x}$$. – Ubiquitous Nov 27 '18 at 12:38

The above answer needs to be supplemented with this answer. You can't just say "the graph looks concave". You have to put in a little work, at least if you were my student.

Let $$x> y$$ without loss of generality (the objective function is symmetrical). Then we have $$U(x,y) = y^2+x$$. Since this is increasing in both variables, we can see right away that our constraint ($$x+py \leq M$$) will bind (you can provide an epsilon delta argument for this if you want). Thus we have $$U(x(y),y) = y^2+M-py$$. This is a polynomial, and we find that $$y = 2/p$$ is the only critical point and by the second derivative test it is a local minimum.

Now we have to consider the case $$x=y$$. This is what the answer above me can be inserted.

Have a good day. Oh and $$x=y$$ is most definitely an interior point in this context. Although technically that solution is on the boundary of the choice set. The only reason it is a special case is that the function isn't differentiable there.