# Another Related Rates Problem

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 0.001$$h^2$$ cubic meters/min, where h is the depth of the water in meters. Will the tank ever overflow?
Thoughts:

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (\frac{4}{5}h)^{2}h$$

$$\frac{dV}{dt}=\frac{16}{25}\pi h^{2}\frac{dh}{dt}$$

Now I replace $$\frac{dV}{dt}$$ with 0.1-0.001$$h^2$$. This is where I am stuck. Any suggestions?

Thanks.

This just gives a separable diff equ....

It can be solved but why solve it ? :

If the water reaches a highest point this means that h(t) has a maximum....

dh/dt=0->h=10m....which curiously is exactly the height of the cone

Is it a maximum ? It turns out the second derivative is zero, the third derivative is zero...aso..

So does it overflow ?

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