As noted above, the fill behaviour is better modeled using Bernoulli's equation. For this problem, we choose point 1 to be at the surface of the water, and point 2 to be at the drain. Bernoulli's equation for this situation is as follows:
P1/(rho*g) + v1^2/(2g) + h1 = P2/rho*g + v2^2/(2g) + h2
P1, P2 = pressures
rho = density of water
g = acceleration due to gravity
v1, v2 = velocities
h1, h2 = heights
To simply the above equation, we make a few assumptions:
(1) The drain is a free jet (i.e. the water surface at the drainage point is open to the atmosphere), so P1 = P2.
(2) The velocity of the water at the surface of the tub is negligible compared to the drainage velocity, so v1 = 0.
(3) We arbitrarily set h2 = 0.
Using these simplifying assumptions, we are left with:
h1 = v2^2/(2g)
Solving for v2, we have:
v2 = SQRT(2g*h1)
We make one more simplifying assumption here, that the cross-sectional area of the tub B is constant all the way up. In that case, V = B*h1. Subbing in, we have:
v2 = SQRT(2g*B*V)
Now, the product of the drainage velocity and the cross-sectional area of the drain Ac gives us the rate of change in volume due to drainage D. Carrying out this operation, we have:
D = Ac*v2 = Ac*SQRT(2g*B*V)
Lumping all the constants into one constant k, we are left with:
D = k*SQRT(V)
We know it takes 20 minutes to drain a full tub, so using a mass balance we have:
dV/dt = -D = -k*SQRT(V)
dV/(-k*SQRT(V)) = dt
Integrating the above expression, with the limits of integration being V = 1...0 (as the tub is being drained) and t = 0...20, we get:
(2/k)*(SQRT(1)-SQRT(0)) = 20-0
2/k = 20
k = 1/10
Now, we know from iamatiger's post that the fill rate F is 1/15. Therefore, the overall balance for the tub is as follows:
dV/dt = F - D = 1/15 - (1/10)*SQRT(V)
This equation is surprisingly involved to tackle, but luckily the Wolfram Alpha site eats equations like this for breakfast!
Here is a link to a clear version of the final answer:
However, a simple asymptote test will gives us some valuable information. Setting dV/dt = 0, we have:
0 = 1/15 - (1/10)*SQRT(V)
SQRT(V) = 2/3
V = 4/9
Therefore the graph of the water volume over time becomes asymptotic as V approaches 4/9, which means that the water will never fill the tub beyond 4/9 of the total volume.
The tub will never fill with the taps on and the drain open. Neat!