Optimization question - water in a conical tank

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SUMMARY

The discussion focuses on optimizing the calculation of water depth in an inverted conical tank with a top radius of 20m and a depth of 15m. The water inflow rate is 0.1m³/min, while the outflow rate is dependent on the water height, calculated as 0.1h³/min. For part (a), the depth increase rate when the water depth is 5m is determined to be approximately 7.1697 x 10-4 m/min. The user seeks clarification on parts (b) and (c) regarding the outflow rate and maximum tank capacity.

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Homework Statement



A water tank is in the shape of an inverted conical cone with top radius of 20m and

depth of 15m. Water is flowing into the tank at a rate of 0.1m^3/min.

(a) How fast is the depth of water in the tank increasing when the depth is 5m?

Water is now leaking from the tank at a rate that depends on the depth h, (h= height of

water in the tank) this rate is 0.1h^3/min.

(b) How fast is the depth of water in the tank changing when the depth is 5m?

(c) How full can the tank get?



Homework Equations





The Attempt at a Solution



Ok for part A:

Tan(angle) = 20/15 = 3/4

So i got a formula for r... r = 4/3 h

Which i put into the formula for the volume of a cone and got:

V = 16pi/27 * h^3

and then differentiated V with respect to time.

dV/dt = 16pi/27 * (3h^2)dh/dt

and i know h and dV/dt so i subbed those in and got

dh/dt = 7.1697 * 10^(-4) m/min


which I'm pretty sure is right.




Now I'm not sure how to do part b.

Do I set dV/dt as 0.1 - 0.1h^3

and then just do the same thing? Or is this wrong. Also, what about part c?

I tend to get a bit lost with these sorts of wordy questions. Please help.
 
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(a) Correct
(b) Yes, that's the right idea.
(c) If you're still stuck after completing (b), post again.
 

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