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

where:

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! ðŸ˜‰ Here is a link to a clear version of the final answer:

http://www.wolframalpha.com/input/?i=dV/dt+%3D+1/15+-+(1/10)*SQRT(V)

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.

**FINAL ANSWER**

The tub will never fill with the taps on and the drain open. Neat!