Is there enough information given to solve this related rates problem? This is the question from a practice exam:
Suppose a pyramid has 4 lateral faces that are all equilateral triangles. Find the rate at which the volume of the pyramid is changing if each side of each triangle increases at a rate of 2 inches per second. Hint: The volume of a square pyramid is given by the formula $V={1\over 3}Bh$, where $B$ is the area of the base of the pyramid and $h$ is the height of the pyramid.
This is the only information I am given. The answer key has just been released, so I know the answer is $16\sqrt{2}in^3/sec$, but I don't know how to get there. And then a friend (who was helping me) thought that maybe there isn't enough information. 
Is there enough?
EDIT: I used the answer given by the key and worked out the problem. I believe that the missing information is "Find the rate at which the volume of the pyramid is changing when the side length is 4 inches." When I put $h$ (height) in terms of $s$ (side length), then differentiated implicitly, and plugging in $4$ for $s$ at the end, I found the solution to be ${16\cdot2\over \sqrt{2}}$, which simplifies to $16\sqrt{2}$ by rationalizing the denominator.  
 A: No, there is not enough information, if you assume that the pyramid is interpreted in the usual way, in the sense of having a square base. Then you can express all the information in terms of the side length. Since the the base is a square, and the height is $\frac{s}{\sqrt{2}}$.
Here's a reference on this type of solid. https://en.wikipedia.org/wiki/Square_pyramid
Now you can just use the normal implicit differentiation technique to see that $\frac{dV}{dt} = f(s)\frac{ds}{dt}$ for an appropriate function of $s$. However,  nowhere in the statement of the problem were you told any information at all about the side length, and there doesn't seem to be a way to deduce it. So unless you leave your answer in this form, you can't go forward.
A: $l$ is the side's length and $h$ the pyramid's  height.
Using Pythagoras' theorem it's trivial to show that the height respects:
$(l/2)^2 + (l/2)^2 + h^2 = l^2$
solve for $h= \frac{l}{\sqrt{2}}$
$B = l^2$ so $V = \frac{Bh}{3} = \frac{l^3}{3\sqrt{2}} $
$\frac{dV}{dt} = \frac{l^2}{\sqrt{2}} \frac{dl}{dt}$
So the answer clearly depends on $l$.
This would clearly always be the case.
Just by dimensional analysis V grows as $l^3$ so the rate of change will always grow as $l^2$.
Unless I'm missing something I think this is it.
