Quote (feanur @ Mar 23 2014 07:07am)
We may assume :
- a constant 350 m^3 per day
- square-base pyramid
- side length 230 metres at the ground
- final height 146 metres
- construction goes on such as if an empty pyramid was filled with water.
using feanur's assumptions:
this problem is a beast, and I won't be formatting a nice version or even solving but I can tell you the method that should work:
the shape we are interested in called a "truncated square pyramid" (a square pyramid with the top sliced off):
its volume is given by
V = (h/3)(a^2 + ab + b^2) = 350*t
take the derivitive of that with respect to time
looking at the picture and image it changing heights, you will realize a is constant with respect to time so treat it as such:
(dv/dt) = (1/3)*[(dh/dt)(a^2 + ab + b^2) + h(a{db/dt} + 2b{db/dt})]=350
we know what a is in feanur's assumptions, b and h you can get with a system of 2 equations: the volume equation above, and the equation for the triangle (below). (dh/dt) is the answer to the whole problem, but there is one stumbling block: (db/dt). We have to express b as a function of these other variables.
This is my attempt, I might have messed up, but even if I did, you need to make an argument more or less like this:
(a-b)/a = h/H
(db/dt) = -(a/H)*(dh/dt)
Substitute that back into the (dv/dt) equation replacing (db/dt) with -(a/H)*(dh/dt). Now with a lot of simplifying you will be able to collect all the (dh/dt) and put it on one side of the equation, and have a numeric value on the other side.