If the surface area of a box is 32 and its volume is doubled what is the new surface area?

Question

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.

Answer 1

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.

Answer 2

The volume of a box is the product of the three dimensions, while the surface of any face is the product of two dimensions.

Hence if you enlarge all dimensions by a factor $f$, the volume increases by $f^3$ and the total surface by $f^2$.

You know that $f^3=2$. Conclude.

Answer 3

Hint: The $r:r^2:r^3$ theorem states that if an object is dilated by $r$, then the surface area increases by a factor of $r^2$, and the volume increases by a factor of $r^3$