# Strictly concave with Non negativity constraint

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

$$f:\mathbb{R^n} \rightarrow \mathbb{R}$$

If the problem is unconstrained, that is: $$\underset{x_1,x_2,\dots,x_n}{max} f(x_1,x_2,\dots,x_n)$$ 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?

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.

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.

• @DrT I was missing a square in the definition of the function, I think the function is now strictly concave. – 5xum Aug 1 '18 at 9:51
• I accidentally deleted my previous comment. Appreciate the answer @5xum! – DrT Aug 1 '18 at 9:52