ANSWER
40807 cm³/min
EXPLANATION
The tank has the shape of a cone, with a total height of 9 meters and a diameter of 3.5 m - so the radius, which is half the diameter, is 1.75 m. As we can see, the relationship between the height of the cone and the radius is,
[tex]\frac{r}{h}=\frac{1.75m}{9m}=\frac{7}{36}\Rightarrow r=\frac{7}{36}h[/tex]So the volume of water will be given by,
[tex]V(h)=\frac{1}{3}(\pi r^2)h=\frac{1}{3}\cdot\pi\cdot\frac{7^2}{36^2}h^2\cdot h=\frac{49\pi}{3888}h^3[/tex]Where h is the height of the water (not the tank).
If we derive this equation, we will find the rate at which the volume of water is changing with time,
[tex]\frac{dV}{dt}=\frac{49\pi}{3888}\cdot3h^{3-1}=\frac{49\pi}{3888}\cdot3h^2=\frac{49\pi}{1296}h^2[/tex]We want to know what is the change of volume with respect to time, and this is,
[tex]\frac{dV}{dt}=\frac{dV}{dt}\cdot\frac{dh}{dt}[/tex]Because the height also changes with time. We know that this change is 24 cm per minute when the height of the water in the tank is 1 meter (or 100 cm), so we have,
[tex]\frac{dV}{dt}=\frac{49\pi}{1296}h^2\cdot\frac{dh}{dt}=\frac{49\pi}{1296}\cdot100^2cm^2\cdot\frac{24cm}{1min}\approx28507cm^3/min[/tex]This is the rate at which the water is increasing in the tank. However, we know that there is a leak at a rate of 12300 cm³/min, which means that in fact the water is being pumped into the tank at a rate of,
[tex]28507cm^3/min+12300cm^3/min=40807cm^3/min[/tex]Hence, the water is being pumped into the tank at a rate of 40807 cm³/min, rounded to the nearest whole cm³/min.
