A manufacturer produces pain relieving tablets - Edexcel - A-Level Maths Pure - Question 2 - 2012 - Paper 3

The surface area A of a cylinder is given by: $A = ext{Base Area} + ext{Lateral Area} = 2 ext{π}x^2 + 2 ext{π}xh$ Substituting h from part (a): $A = 2 ext{π}x^2 + 2 ext{π}x \left( \frac{60}{ ext{π}x^2} \right)$ This simplifies to: $A = 2 ext{π}x^2 + \frac{120}{x}$

To find the minimum value of A, we first differentiate A with respect to x: $A' = \frac{dA}{dx} = 4 ext{π}x - \frac{120}{x^2}$ Setting the derivative to zero for critical points: $4 ext{π}x - \frac{120}{x^2} = 0$ Solving this gives: $4 ext{π}x^3 = 120 \implies x^3 = \frac{120}{4 ext{π}} \implies x = \sqrt[3]{\frac{30}{\text{π}}}$

Substituting the value of x back into the surface area formula: $A = 2 ext{π}\left(\sqrt[3]{\frac{30}{\text{π}}}\right)^2 + \frac{120}{\sqrt[3]{\frac{30}{\text{π}}}}$ Calculating this expression will yield the minimum value of A, rounded to the nearest integer.

To confirm that this value of A is a minimum, we check the second derivative: $A'' = \frac{d^2A}{dx^2} = 4 ext{π} + \frac{240}{x^3}$ Since both terms are positive (for x > 0), it implies that A is a minimum at the critical point found in part (c).