A solid right circular cylinder has radius r cm and height h cm - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 2

To find the maximum volume using calculus, we first take the derivative of V with respect to r:

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

Next, we set the derivative equal to zero to find critical points:

$400 - 2\pi r = 0\Rightarrow r = \frac{400}{2\pi} = \frac{200}{\pi}$

Now we evaluate V at this critical point:

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

Calculating this gives:

$V = \frac{80000}{\pi} - \pi \left( \frac{40000}{\pi^2} \right) = \frac{80000}{\pi} - \frac{40000}{\pi} = \frac{40000}{\pi}$

Now let’s calculate this numerically:

$V \approx \frac{40000}{3.14159} \approx 12732.4 \text{ cm}^3$

Thus, rounding to the nearest cm³, the maximum volume is approximately:

$12732 \text{ cm}^3$

To confirm that the value found is indeed a maximum, we must check the sign of the second derivative:

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

Since $-2\pi < 0$, the second derivative is negative, which indicates that the function V is concave down at the critical point. Therefore, the maximum value of V occurs at:

$r = \frac{200}{\pi}$

Thus, the value of V found is indeed a maximum.