Optimum Tapping

Suppose the filled pyramid has a height of $x$ . Since it's a pyramid has a square base and is constructed within a cube, its volume will equal $\dfrac{x^3}{3}$ due to triangular similarity.

Now we can set up the equation for rates of change:

$\dfrac{dv}{dt} = \dfrac{dv}{dh}\cdot \dfrac{dh}{dt}$

We can evaluate $\dfrac{dv}{dh} = \dfrac{d}{dh} \dfrac{x^3}{3} = x^2$ .

For tap $A$ , the rate of change at $h = x+1$ equals: $\dfrac{dv}{dt} = 1080 = (x+1)^2\cdot \dfrac{dh_{a}}{dt}$ .

For tap $C$ , the rate of change at $h = x-1$ equals: $\dfrac{dv}{dt} = 320 = (x-1)^2\cdot \dfrac{dh_{c}}{dt}$ .

The $\dfrac{dh}{dt}$ for tap $A$ is $1.5$ times of that for tap $C$ .

Hence, when we divide the first equation with the second one, we will get:

$\dfrac{1080}{320} = \dfrac{ (x+1)^2}{ (x-1)^2}\cdot 1.5$

$\dfrac{9}{4} = \dfrac{ (x+1)^2}{ (x-1)^2}$

$\dfrac{3}{2} = \dfrac{x+1}{ x-1}$

$3x -3 = 2x + 2$

$x = 5$

Now all we need to know is the time $T$ taken to obtain such volume $V$ , and since the rate of change is constant, we can calculate for tap $A$ :

$\dfrac{dv}{dt} = \dfrac{V}{T} = 1080$

$1080 = \dfrac{6^3}{3T}$

$T = \dfrac{72}{1080} = \dfrac{1}{15}$

Finally, we can calculate the flow rate for tap $B$ :

$\dfrac{dv}{dt} = \dfrac{V}{T} = 15\cdot \dfrac{1}{3}\cdot 5^3 = \boxed{625}$

Therefore, the tap $B$ has constant flow rate of $\boxed{625}$ mL/min.