A steel manufacturing company wants to manufacture an open square-based stainless steel container with a volume of 4 000 cm³ - NSC Technical Mathematics - Question 8 - 2021 - Paper 1

The surface area $S$ of the open square-based container can be expressed using the formula:

$Surface \ Area = length \times breadth + 2 \times length \times height + 2 \times breadth \times height$

Since the length and breadth are both $x$, we substitute:

$S = x \times x + 2 \times x \times h + 2 \times x \times h$

This simplifies to:

$S = x^2 + 4xh$

Now substituting for $h$ using our earlier result:

$S = x^2 + 4x \left(\frac{4000}{x^2}\right)$

Which further simplifies to:

$S = x^2 + \frac{16000}{x}$

To minimize the surface area $S = x^2 + \frac{16000}{x}$, we can find the derivative:

$\frac{dS}{dx} = 2x - \frac{16000}{x^2}$

Setting the derivative equal to zero for critical points gives:

$2x - \frac{16000}{x^2} = 0$

Multiplying through by $x^2$ results in:

$2x^3 = 16000$

Thus:

$x^3 = 8000 \Rightarrow x = 20$

Now substituting back to find $h$:

$h = \frac{4000}{x^2} = \frac{4000}{20^2} = \frac{4000}{400} = 10$

Therefore, the height of the container is $10$ cm.