We prove that the greedy algorithm uses the fewest possible trucks by showing that it “stays ahead” of any other solution. Specifically, we consider any other solution and show
the following. If the greedy algorithm fits boxes b1, b2, . . . , b into the first /c trucks, and the other solution fits b1, . . . , b into the first k trucks, then i < j. Note that this implies the optimality of the greedy algorithm, by setting k to be the number of trucks used by the
greedy algorithm.