Suppose we have to fill our bag with items and each item has a value $V_i \sim \text{Uniform}[a,b]$ where each random variable is iid. There are $N$ items in total but only a fraction $x$ of all items can be put inside the bag. When an item is put inside the bag or discarded the next item appears and you cannot go back. The goal here is to maximize the total value of the bag, i.e., $\max_B \sum_{i\in B}V_i$.
A naive strategy is to put all items in the bag until the bag is full, with corresponding expectation: $E(B)=xN\frac{b-a}{2}$.
My idea is to put all items with values between $[b - y(b-a), b]$ in the bag, where $y$ is updated, i.e., $y=$ number of remaining items in bag divided by$ N$. However I don't know what the expectation of this strategy is.
What is the best strategy for this problem? The problem is somewhat similar to the secretary problem but now we are looking for the best $x$ fraction of secretaries, instead of the best single secretary.
Let's denote by $c_{m,n}$ the expected value of the value we get following a best policy when there's room $m$ in bag and there are $n$ items to come. This is the "future value", it doesn't consider what we have in our bag right now. The process kind of starts a new every time and only thing that matters is the room in our bag and how many items are to come.
This should hold:
$$ c_{m,n} = \mathbb{E}[\max (V+c_{m-1, n-1}, c_{m, n-1})], \\ c_{0,n} = 0 \text{ and } c_{m,0} = 0 $$
Writing the expectation we get
$$ c_{m,n} = \frac{1}{b-a}\int_a^b \max(v+c_{m-1, n-1}, c_{m, n-1}) dv $$
I don't know if we can arrive at a closed formula, but these can be tabulated. For example for $a=0, b=1$ and $0\leq m,n \leq 8$
$$ \displaystyle \left(\begin{array}{rrrrrrrrr} 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 & 0.0000 \\ 0.0000 & 0.5000 & 0.6250 & 0.6953 & 0.7417 & 0.7751 & 0.8004 & 0.8203 & 0.8364 \\ 0.0000 & 0.5000 & 1.000 & 1.195 & 1.320 & 1.409 & 1.476 & 1.529 & 1.571 \\ 0.0000 & 0.5000 & 1.000 & 1.500 & 1.742 & 1.909 & 2.034 & 2.132 & 2.211 \\ 0.0000 & 0.5000 & 1.000 & 1.500 & 2.000 & 2.275 & 2.476 & 2.632 & 2.757 \\ 0.0000 & 0.5000 & 1.000 & 1.500 & 2.000 & 2.500 & 2.800 & 3.029 & 3.211 \\ 0.0000 & 0.5000 & 1.000 & 1.500 & 2.000 & 2.500 & 3.000 & 3.320 & 3.571 \\ 0.0000 & 0.5000 & 1.000 & 1.500 & 2.000 & 2.500 & 3.000 & 3.500 & 3.836 \\ 0.0000 & 0.5000 & 1.000 & 1.500 & 2.000 & 2.500 & 3.000 & 3.500 & 4.000 \end{array}\right) $$