Let $h(t)$ be the height of the water at time $t$, and let $r(t)$ be the radius of the surface of the water at time $t$; from similar triangles we know that
$$\frac{h(t)}{r(t)}=\frac{14}{2.75}=\frac{56}{11}\;.$$
We could solve this for either $r(t)$ or $h(t)$ in terms of the other, but notice that we’re told $h'(t)$ at a particular moment, and we’re not told anything about an specific value of $r(t)$. This suggests that we’d be better off working in terms of $h(t)$, so we’ll solve for $r(t)$ and get $$r(t)=\frac{11}{56}h(t)\;.$$
At time $t$ the volume $V(t)$ of water in the tank is the volume of a right circular cone with height $h(t)$ and base radius $r(t)$, which is given by
$$V(t)=\frac13\pi r(t)^2h(t)=\frac{\pi}3\left(\frac{11}{56}h(t)\right)^2h(t)=\frac{121\pi}{9408}h(t)^3\;.$$
Then
$$V'(t)=\frac{121\pi}{3136}h(t)^2h'(t)\;.$$
We’re told that $h'(t)=0.24$ when $h(t)=3$; if we call that moment time $t_0$, we have
$$V'(t_0)=\frac{121\pi}{3136}\cdot3^2\cdot0.24=\frac{3267\pi}{39200}\text{ m}^3/\text{min}\;.$$
Now let $v$ be the rate in cubic metres per minute at which water is being pumped into the tank. Taking into account both the inflow and the leakage, we know that at all times
$$V'(t)=v-0.0068\text{ m}^3/\text{min}\;.$$
In particular, at time $t_0$ we have
$$v-0.0068=\frac{3267\pi}{39200}\;,$$
which is completely straightforward to solve for $v$.
The volume is given by:
$V=\frac{1}{3}\pi r^2h$
You are given $\frac{dV}{dt}$, so you will have to take the derivative of the volume function with respect to time. Keep in mind that the radius and the height are NOT constants, they are variables. However, they are proportional:
$\frac{h_1}{r_1}=\frac{h_2}{r_2}$
This is because, if you draw a triangle straight down in the cone, you'll get similar triangles, as I, a true artist, have demonstrated below:
This is true for RIGHT cones.
We are given the cone's actual height and radius, and so we know that:
$\frac{h}{r}=\frac{20}{10}=2$
Hence:
$r=\frac{h}{2}$
I found what r is because I want to use it in the equation because we are seeking for the change in the height over time. Let's plug it into the original equation, then:
$V=\frac{1}{3}\pi (\frac{h}{2})^2h=\frac{1}{12}\pi h^3$
Take the derivative of this with respect to time:
$\frac{dV}{dt}=\frac{1}{4}\pi h^2 * \frac{dh}{dt}$
This by the way always happens, we just never see it. Take for instance:
$y= 5x^2+x$
When you differentiate, the answer is really:
$\frac{dy}{dx}=10x*\frac{dx}{dx}+\frac{dx}{dx}$
But $\frac{dx}{dx} = 1$ so we never show it.
Back to what I was saying, the next step is to determine $\frac{dh}{dt}$, which is what we are trying to solve. So we need the instantaneous height of the cone when it's 1/8 filled. 1/8 filled means the cone's volume is 1/8 of its original volume. Its original volume is (and you can use $V=\frac{1}{3}\pi r^2h$, but I find the function of volume against height only easier since there's only 1 variable):
$V=\frac{1}{12}\pi h^3$
$V_{full}=\frac{1}{12}\pi (20)^3 =$ some value I stored on my calculator.
Multiply that by 1/8 and you get the volume of the tank at that time. Use the equation to find the instantaneous height:
$V_{instantaneous}=\frac{1}{12}\pi h^3$
The height is 10. You can now plug it into the derivative equation we had:
$\frac{dV}{dt}=\frac{1}{4}\pi h^2 * \frac{dh}{dt}$
We are given that $\frac{dV}{dt}=2$, and we now know the instantaneous height. The answer is then:
$\frac{dh}{dt}=\frac{2}{25 \pi} \approx 0.025 \frac{ft}{min}$
Let me know if I made any mistakes before you down vote me into oblivion.
Best Answer
Let $h$ denote the height, and $r$ the radius of the part of the cone that is filled with water. Since the total height is 6 and the radius at the top is 2, using proportions we have $r=\frac h3$. The volume of cone is $V=\pi r^2 h/3 = \pi h^3/27$. Here both $V$ and $h$ are functions of the time $t$, so differentiating both sides with respect to $t$ we get $V'=3h^2\pi h'/27=h^2\pi h'/9$. It is given that $h'=20$ when $h=2 m = 200 cm$, so we have $V'= 200^2\cdot 20\pi /9 \approx 279252.68$. Adding the 9500 we get $288752.68\approx 288753$. (It appears to me you got the answer mostly right, except the "Round your answer to the nearest integer" part.)