If I chop my apple into smaller pieces, can I fit more apple in my lunchbox? Well, that's how my little sister asked it, and I couldn't provide a quick answer.
If we ignore any changes in shape from the cutting process, then my first thought was that the scaling wouldn't make any difference. But then I started thinking that more efficient packings might only become possible after we cut the apple pieces down to a certain size.
I realise this is almost unanswerable without multiple assumptions (spherical pieces of apple anyone? :) ) but at the end of the day I'm interested in what is known (if anything) about the packing of irregularly shaped solids in finite containers of similar dimension (within an order of magnitude) to the solids themselves. 
Thanks
 A: This restates some of the discussion in the comments, but here is a more rigorous proof of what I take this question to be asking:
Say that a 3D shape is "reasonable" if it has positive volume and the limit as $\epsilon\to0^+$ of the volume of the points within distance $\epsilon$ of its boundary is $0$.
In apple terms, we're requiring the apple is not all peel - if you take a fine enough apple-peeler, you can remove the peel without taking out any significant part of the apple. (This is to avoid pathological shapes that are arbitrarily close to their boundary and otherwise hard to dice up nicely.)
Claim: Given a reasonably-shaped apple and a reasonably-shaped lunchbox with more volume than the apple, it is possible to cut the apple into finitely many pieces with a knife so that they can be rearranged to fit inside the lunchbox.
Proof: Suppose that the ratio $\frac{\text{Volume}(\text{box})}{\text{Volume}(\text{apple})}$ is equal to some $r>1$.
Let $\epsilon_1>0$, and cut the apple along planes normal to the $x$, $y$, and $z$ axes so that it is cut along a grid of cubes of side length $\epsilon_1$. Extend every nonempty piece until it fills up its corresponding cube. By our reasonableness property, this will give us a set of cubes with total volume approaching $\text{Vol}(\text{apple})$ from above as $\epsilon_1\to0$. (This is because any cube which contains some of the apple has no points more than $\epsilon_1\sqrt{3}$ outside the boundary.)
Conversely, if we cut the box along planes normal to the $x$, $y$, and $z$ axes so that it is cut along a grid of cubes of side length $\epsilon_2>0$, our reasonableness property tells us that the cubes which fit entirely inside the box (without touching the boundary) will have total volume approaching $\text{Vol}(\text{box})$ from below as $\epsilon_2\to0$.
So then we can choose $\epsilon$ small enough that cutting the box into cubes of side length $\epsilon$ gives us "inside cubes" with total volume at least $\frac{\text{Vol}(\text{box})}{\sqrt{r}}$, and cutting the apple into cubes of side length $\epsilon$ gives us "outside cubes" with total volume at most $\text{Vol}(\text{apple})\cdot\sqrt{r}$.
But then we have at least as many empty lunchbox cubes to fill as we have partially-filled apple cubes to place inside, so we can just pair up the (identically-shaped) cubes with each other however we like. This completes the proof. (In fact, for most nice apple and lunchbox shapes it will do so in a relatively small number of knife cuts, because we only need to make $O(1/\epsilon)$ cuts in order to produce $O(1/\epsilon^3)$ pieces.)

If you place any absolute bound on the number of pieces, then it is not necessarily possible to do this, even for apples and lunchboxes of similar dimensions. Consider a worm-eaten apple with many tiny holes inside it, so that it is more like a collection of soap bubble films - even though its volume is very small, it takes a large number of cuts to allow these holes to be filled nicely and to pack it into a tighter space.
