‘n Koeldrankblikkie het ‘n volume van 340 cm³, ‘n hoogte van h cm en ‘n radius van r cm - NSC Mathematics - Question 9 - 2016 - Paper 1

The surface area A of the cylinder consists of the area of the top and bottom circles and the lateral surface area:

$A = 2\pi r^2 + 2\pi rh$

By substituting the expression found in 9.1 for h, we have:

$A = 2\pi r^2 + 2\pi r \left(\frac{340}{\pi r^2}\right)$

Simplifying this:

$A(r) = 2\pi r^2 + 680\cdot \frac{1}{r}$

Following this, we can simplify to the form given in the question.

To find the radius that minimizes the surface area, we need to derive A with respect to r and set the derivative to zero:

$A'(r) = 4\pi r - 680 r^{-2}$

Setting the derivative to zero for minimization:

$4\pi r - \frac{680}{r^2} = 0$

Rearranging gives:

$4\pi r = \frac{680}{r^2}$

Multiplying through by r² leads to:

$4\pi r^3 = 680$

Solving for r, we have:

$r^3 = \frac{680}{4\pi}$

Taking the cube root:

$r = \sqrt[3]{\frac{680}{4\pi}} \\ r \approx 3.78 ext{ cm}$