A company has been contracted to manufacture right cylindrical cans to package baked beans - NSC Technical Mathematics - Question 8 - 2023 - Paper 1

The total surface area $A$ of the cylindrical can is given by:

$A = 2 \times (area \ of \ the \ base) + (perimeter \ of \ the \ base) \times height$

The area of the base is $\pi r^{2}$ and the perimeter of the base is $2\pi r$. Thus, substituting these into the equation:

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

Now, we substitute $h$ from part 8.1:

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

This simplifies to:

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

To find the minimum total surface area, we first derive the surface area function $A(r)$:

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

Taking the derivative, we have:

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

Setting this derivative equal to zero for minimization gives:

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

Rearranging leads to:

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

Solving for $r$:

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

$r = \sqrt[3]{\frac{700}{4\pi}} \approx 3.82 \ ext{ cm}.$

Now, substituting $r$ back into the height formula from part 8.1:

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

Therefore, the dimensions of the can are approximately:

• Radius: $3.82$ cm
• Height: $7.63$ cm.