Question

Optimisation help please

Answer 1

This is how I've worked it:
r = h/4, so plugging that into the volume formula, v=(1/3)(pi)(r^2)(h), I get v = (pi)(h^3)/48. So then, dv/dt = (pi)(h^2)/16, right? But after this, I am confused how to proceed.

Answer 2

@FibonacciChick666 @zepdrix

Answer 3

Mmm still trying to work this one out myself >.<
But I did notice... umm

we're taking our derivative with respect to `time`.
So don't forget to chain rule on your h.
You should be getting a h' (dh/dt) popping out, yes?

Answer 4

\[\Large\rm V=\frac{1}{48}\pi h^3\]\[\Large\rm V'=\frac{1}{16}\pi h^2h'\]

Answer 5

And then I think what we have going on is like...
This V' represents the `net change` of the volume.
\[\Large\rm V'=V_{out}+V_{in}\]giving us,\[\Large\rm V_{out}+V_{in}=\frac{1}{16}\pi h^2h'\]They told us the rate at which water is coming in.
Plugging all that information in should allow us to figure out at what rate we're leaking.

Answer 6

Thanks, that helped, I got it; I was just starting to think along those lines. xD

Answer 7

I forgot the dh/dt bit.

Answer 8

cool c: