Can anyone help?

[ziddy83]

I need help with the following problem...

The volume (in gallons) of water in a tank at time t seconds is given by the function
V(t) = e^{2t} - 12t^2+100 where 0 \leq t \leq 3

a) when is the water flowing out of the tank at the fastest rate? At what rate is it flowing at this time?

b) When is the water flowing into the tank at the fastest rate? At what rate is it flowing at this time.

So to start this, will i take the derivative of the volume function and then plug in a value to find time T? :confused:

[Justin Lazear]

Water is flowing in most quickly when the volume functions is increasing most quickly. So you'd need to maximize the rate of change of the volume, i.e. dV/dt. Water is flowing out most quickly when the volume function is decreasing most quickly, so you'd need to minimize the rate of change of the volume.

So first you differentiate V(t), then it's a simple maximization/minimization problem using y(t) = dV/dt. Maximize and minimize y(t).

--J

[ziddy83]

cool, thanks

[ziddy83]

Can someone help me with the max/min part? After i take the derivative...do i then plug in the two values i have been given (0 and 3?) Im a little confused on that part, thanks.

[thermodynamicaldude]

The question is asking for "when", which implies time. THus, take the derivatives, and solve for t when the derivative = 0. AFter you have found your values for t, then you will need to test whether the value is a maximum or a minimum.

(if you don't find a max or min. with these values, then the max or min is porbably either one of the endpoints (i.e, when t = 0 or 3)

Keep in mind that the values for t must be between 0 and 3.

[ziddy83]

so by simply plugging in the two values in the dv/dt function, i should have my max and min rate? Because when i plug in 3 to the V' function i get 2348.6, and thats...the rate of the water, but how do i know thats the max rate?

[ziddy83]

oh ok, you just edited your post, thanks

[ziddy83]

YES i think i got it...thanks to everyone who helped.