A manufacturer produces pain relieving tablets - Edexcel - A-Level Maths Pure - Question 8 - 2012 - Paper 3

The surface area A of a cylinder is given by:

$A = 2 ext{π}r^2 + 2 ext{π}rh$

Substituting r with x, and h from part (a):

$A = 2 ext{π}x^2 + 2 ext{π}x \left( \frac{60}{ ext{π}x^2} \right)$

Simplifying this:

$A = 2 ext{π}x^2 + \frac{120}{x}$

To find the minimum value of A, we first find the derivative A':

$A' = \frac{d}{dx}(2 ext{π}x^2 + \frac{120}{x})$

Calculating the derivative gives:

$A' = 4 ext{π}x - \frac{120}{x^2}$

Setting A' to 0 for minima:

$4 ext{π}x - \frac{120}{x^2} = 0$

Rearranging gives:

$4 ext{π}x^3 = 120$

So aftre simplifying:

d $$ x = \sqrt[3]{\frac{30}{ ext{π}}}$$

Using the value of x found in part (c), we substitute it back into the surface area formula:

$A = 2 ext{π} \left( \sqrt[3]{\frac{30}{ ext{π}}} \right)^2 + \frac{120}{\sqrt[3]{\frac{30}{ ext{π}}}}$

Calculating this value will give the minimum surface area. To find the numerical value, it may be computed using a calculator and should be rounded to the nearest integer.

To verify that the value of A is a minimum, we must check the second derivative, A'':

$A'' = \frac{d^2}{dx^2}(2 ext{π}x^2 + \frac{120}{x}) = 4 ext{π} + \frac{240}{x^3}$

Since both terms are positive, A'' > 0 indicates that A has a local minimum at the x found in part (c). Hence, the value of A calculated in part (d) is indeed a minimum.