## Related Rates and a Conical Tank

A water tank is in the shape of an inverted cone with depth 10 meters and top radius 8 meters. Water is flowing into the tank at 0.1 cubic meters/min but leaking out at a rate of 0.001h2 cubic meters/min, where h is the depth of the water in the tank in meters. Can the tank ever overflow?

Can anyone help with this? The flowing in and leaking out is a bit confusing.

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 $$V = \frac{1}{3}\pi r^{2} h$$ You know that $$\frac{r}{h} = \frac{4}{5}$$, so $$r = \frac{4h}{5}$$ So the new expression is: $$V = \frac{1}{3}\pi (\frac{4h}{5})^{2}h = \frac{16\pi}{75}h^{3}$$ You also know that $$\frac{dV}{dt} = 0.1 - 0.001$$