Strictly concave with Non negativity constraint

Question

Say I intend to maximize a real-valued, strictly concave, twice-differentiable function:

\begin{equation} f:\mathbb{R^n} \rightarrow \mathbb{R} \end{equation}

If the problem is unconstrained, that is: \begin{equation} \underset{x_1,x_2,\dots,x_n}{max} f(x_1,x_2,\dots,x_n) \end{equation} I would take the first derivative, and set equal to zero. This would yield the unique global optimum due to the strict concavity of the problem.

Say I impose a non-negativity constraint, such that my problem becomes:

\begin{aligned} & \underset{x_1,x_2,\dots,x_n}{max} & f(x_1,x_2,\dots,x_n) \\ & & x_1,x_2,\dots,x_n \geq 0 \end{aligned}

Will it be sufficient to take the solution to the unconstrained problem and set any negative $x_i$'s (if any) equal to 0? Or does that depend entirely on what my function look like, regardless of the real-value, strictly concave, and twice-differentiable attributes of it?

In other words, can I bypass using Kuhn-Tucker conditions due to the nature of the function?

Answer 1

Will it be sufficient to take the solution to the unconstrained problem and set any negative $x_i$'s (if any) equal to $0$?

NO.

the function

$$f(x,y) = -(x+y+1)^2 - (x-2y+2)^2$$

is a counterexample. The global maximum of the function is clearly at $x=\frac43, y=-\frac13$, where the otherwise negative function hits $0$. Proving that the function is positive definite should also be fairly simple.

However, the point $(\frac43, 0)$ is not the maximum of the function on $\{x\geq 0,y\geq 0\}$, since $f(\frac43, 0) = -16.\dot 5$ while $f(0, 0.6) = -3.2$ which is a greater value.

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Naturally, if all the $x_i$ of the global maximum are positive, then yes, your global maximum is also a maximum on $\mathbb R_+^n$. However, if one of the values is negative, you cannot simply project your point the way you suggest.