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Suppose the utility function is continuous, differentiable, strictly increasing and strictly quasiconcave. Whether the utility maximization problem has unique interior solution? If not, is there any counterexample?

My idea: We know that the utility maximization problem has unique solution when the utility function is strictly quasiconcave. And we know that corner solution usually occurs when the utility function is quasilinear or $\min\{x_1,x_2\}$, and these cases have already been ruled out by the strictly quasiconcavity of $u(\cdot)$. Intuitively, I think it should has no corner solution by observing the graphs, but I don't know how to prove it precisely.

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  • $\begingroup$ Just a minor quibble: interior solutions are not the antonym of corner solutions. Rather, an interior solution to an optimization problem is one that lies in the interior of the domain (i.e. a solution that lies in the union of all open sets contained in the domain). With increasing utilities solutions always lie on the boundary (more specifically,. budget constraints bind). $\endgroup$ Oct 12, 2023 at 5:51

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Consider a consumer with utility function $u:\mathbb{R}_{++}\times\mathbb{R}_+\rightarrow\mathbb{R}$ defined as follows: $u(x,y)=\sqrt{x}+y$. It is continuous, differentiable, strictly increasing, strictly quasi-concave, but there are possibilities when there is a corner solution to the utility maximisation problem of such a consumer. Please see this for details: https://economics.stackexchange.com/a/16475/11824 Above was the case of quasi-linear utility which is a counterexample to your claim.

Another counterexample is of a consumer with utility function $u:\mathbb{R}^2_+\rightarrow\mathbb{R}$ defined as follows: $u(x,y) = (x+1)(y+1)$. This is also continuous, differentiable, strictly increasing, strictly quasi-concave utility function, but there are possibilities of a corner solution when we solve the utility maximisation problem of this consumer. For example: Consider the utility maximisation problem: \begin{eqnarray*} \max_{x\geq 0, \ y\geq 0} & (x+1)(y+1) \\ \text{s.t. } & 3x+y\leq 1\end{eqnarray*} Solving this problem, we get $(x^d,y^d)=(0,1)$ which is in the corner.

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