Convex function subject to $0\le x_1\le \ldots \le x_n\le 1$ and linear constraint

I am maximizing a convex function (a positive definite quadratic form, if it makes a difference) subject to $$0\le x_1\le \ldots \le x_n\le 1$$ and a linear constraint $$a^\top x+b=0$$. Can I conclude that the maximum occurs where some of those inequalities are equalities, or something of the sort?

Yes. Your feasible region is a convex (bounded) polytope, which means that every point in the feasible region can be written as a convex combination of extreme points. Let's say that your function is $$f()$$, the maximum occurs at $$\bar{x}\in [0,1]^n$$, and $$\bar{x} = \sum_{i=1}^k \alpha_i x^{(i)}$$, where the $$x^{(i)}\in [0,1]^n$$ are extreme points of the feasible region and the $$\alpha_i$$ are nonnegative weights with $$\sum_{i=1}^k \alpha_i = 1$$. By the convexity of $$f()$$, $$f(\bar{x}) \le \sum_{i=1}^k \alpha_i f(x^{(i)}),$$so for $$\bar{x}$$ to maximize $$f()$$ over the feasible region every $$x^{(i)}$$ in the sum with $$\alpha_i >0$$ must maximize $$f()$$.
An extreme point (corner) of a polytope is the intersection of a bunch of linearly independent bounding hyperplanes (binding constraints), so there will be at least one maximizer of $$f()$$ where $$n$$ of the constraints are binding.
• When you say at least $n$ of the constraints are binding at the corner, $n$ is the number of hyperplanes/constraints that intersect to make up that corner of the polytope? Sep 10, 2021 at 2:15