TITLE: If the surface area of a box is 32 and its volume is doubled what is the new surface area?
QUESTION [0 upvotes]: Original surface area :32
Original volume: x
New volume: 2x
What is the new surface area? Please provide an explanation or show work, I don't know how to do it.

REPLY [0 votes]: It is not too hard (Lagrange multipliers) to show that for a given volume $x$,
the possible outside surface areas $S$ of a (closed) box satisfy
$S \ge 6 \sqrt[3]{x^2}$.
We are given $32 \ge 6 \sqrt[3]{x^2}$, so we have $x^2 \le ({ 32 \over 6})^3$,
that is, an upper bound on the volume.
If we double the volume to $2x$, we see that the allowable surface areas
satisfy $S \ge 6 \sqrt[3]{4x^2}$.
Since all we have is an upper bound on $x$, we cannot say very much.
Perhaps the best we can say is that we can guarantee that any surface
area $S \ge 32 \sqrt[3]{4}$ is achievable.
{"set_name": "stack_exchange", "score": 0, "question_id": 1293412}