The problem consists in the standard knapsack problem in interger programming with the weights that all have the same values, for example they are all equal to one.
It seems to me that the solution of this problem should be trivial, that is first set to one the variable with the highest profit, then the one with the second highest profit and so on, until the knapsack is full.
I was wondering if there is a formal way to prove that this procedure reaches a global minimum, or if there are papers that deal with this particular instance of the knapsack problem.
Thank you.
Try proof by contradiction. Assume all weights are 1 and the knapsack capacity is $C$. Posit an optimal solution that includes at least one item not among the $C$ most valuable items, and prove that you can improve it (contradicting optimality).