A solid right circular cylinder has radius r cm and height h cm - Edexcel - A-Level Maths Pure - Question 3 - 2009 - Paper 2

To find the maximum value of V, we differentiate V with respect to r:

$\frac{dV}{dr} = 400 - 2\pi r$

Setting the derivative equal to zero to find critical points:

$400 - 2\pi r = 0$

Solving for r gives:

$r = \frac{400}{2\pi} = \frac{200}{\pi} \approx 63.66 \text{ cm}$

Now we find the second derivative to confirm it's a maximum:

$\frac{d^2V}{dr^2} = -2\pi$

Since this is negative, it confirms a maximum. Now we substitute $r = \frac{200}{\pi}$ back into the volume equation:

$V = 400(\frac{200}{\pi}) - \pi (\frac{200}{\pi})^2$

Evaluating:

$V = \frac{80000}{\pi} - \frac{40000}{\pi} = \frac{40000}{\pi} \approx 12731 \text{ cm}^3$

So, the maximum volume to the nearest cm³ is 12731 cm³.

We have already confirmed that the second derivative $\frac{d^2V}{dr^2} = -2\pi$ is negative, which indicates concavity downwards. This means that the critical point found at $r = \frac{200}{\pi}$ is indeed a maximum.

Furthermore, since we worked through the calculus of the function and identified that $\frac{dV}{dr}$ changes from positive to negative around this critical point, it provides further confirmation that the value of V found is truly a maximum.