1
$\begingroup$

The volume of water (litres) which has flowed through a swimming pool filter t minutes after starting it is $$V=\frac{1}{100}(30t^3-\frac{t^4}{4})$$ where $$0 \le t \le 90$$

when is the greatest rate of flow?

I am a little bit confused with this question because when I differentiate and set it to $0$, I get two roots $t=0$ and $t=90$. However, the answer tells me that the greatest rate of flow occurs when $t=60$

I've verified that my differentation is correct so I'm not sure how to get the maximum, is there something that I am missing?

  • 0
    how many times did you differentiate it? When you differentiate to get dV/dt - what is dV/dt? What is V measuring? – 2017-02-13
  • 0
    you are missing something, yes - can you see what V is, and then what it is you want to find and maximize? – 2017-02-13
  • 0
    @mannav - sorry that isn't right, an important fact is missing here. – 2017-02-13
  • 0
    V measures one quantity and dV/dt is a measurement of another quantity. – 2017-02-13

1 Answers 1

2

By setting $\frac{dV}{dt}$ to $0$, you are finding the extrema of $V$, not of the rate.

What you need is to maximize $\frac{dV}{dt}$, for which the extremum can be found by setting the derivative of $\frac{dV}{dt}$ to $0$.

So, $$\frac{d^2V}{dt^2}=\frac{1}{100}(180t-3t^2)=0$$ which gives $$180t=3t^2$$ $$t=0,60$$

Taking the second derivative of $\frac{dV}{dt}$, using the second derivative test, $$\frac{d^3V}{dt^3}(0)=\frac{9}{5}\gt0\implies\text{minimum}$$ $$\frac{d^3V}{dt^3}(60)=-\frac{9}{5}\lt0\implies\text{maximum}$$

Thus, the maximum rate of flow is at $t=60$.

  • 0
    Good answer +1, the trick here of course is knowing that it is the *rate of flow of volume* that is to be maximised, hence effectively solutions to $\ddot{V}=0$ are required. – 2017-02-13
  • 0
    It is worth pointing out that there isn't an absolute max at $t=90$. – 2017-02-13