'n Maatskappy is gekontrakteer om regte silindervormige blikkies te vervaardig om sousbone te verpakk - NSC Technical Mathematics - Question 8 - 2023 - Paper 1

First, we express the total surface area $A$ as follows:

$A(r) = 2\pi r^{2} + 2\pi r h$

Substituting the expression found for $h$:

$A(r) = 2\pi r^{2} + 2\pi r \left(\frac{350}{\pi r^{2}}\right)$

Simplifying gives:

$A(r) = 2\pi r^{2} + \frac{700}{r}$.

To find the dimensions when the total surface area $A(r)$ is at its minimum, we need to take the derivative of $A(r)$:

$A'(r) = 4\pi r - \frac{700}{r^{2}}$

Setting the derivative to zero for critical points:

$4\pi r - \frac{700}{r^{2}} = 0$

Solving for $r$ gives:

$4\pi r^{3} = 700$

$r^{3} = \frac{700}{4\pi} \implies r = \sqrt[3]{\frac{700}{4\pi}} \approx 3.82 \text{ cm}$

Then substituting to find $h$:

$h = \frac{350}{\pi (3.82)^{2}} \approx 7.63 \text{ cm}$