Figure 4 shows an open-topped water tank, in the shape of a cuboid, which is made of sheet metal - Edexcel - A-Level Maths Pure - Question 1 - 2007 - Paper 2

To find the stationary points, we differentiate A with respect to x and set the derivative to zero:

$A = \frac{300}{x} + 2x^2$

Differentiating gives: $\frac{dA}{dx} = -\frac{300}{x^2} + 4x.$

Setting the derivative to zero for stationary points results in: $-\frac{300}{x^2} + 4x = 0$

Multiplying through by $x^2$ yields: $-300 + 4x^3 = 0$

Thus: $$4x^3 = 300 \Rightarrow x^3 = 75 \Rightarrow x = \sqrt[3]{75} \approx 4.22.$

Therefore, the value of x for which A is stationary is approximately 4.22.

To confirm that the calculated value of x gives a minimum value, we can compute the second derivative:

$\frac{d^2A}{dx^2} = \frac{600}{x^3} + 4.$

Given that both terms are positive for $x > 0$, the second derivative is positive, indicating that A has a local minimum when x is approximately 4.22.

Substituting $x = \sqrt[3]{75}$ back into the area formula:

$A = \frac{300}{x} + 2x^2.$

Calculating with $x \approx 4.22$ gives:

$A \approx \frac{300}{4.22} + 2(4.22)^2 \approx 71.02 + 35.57 \approx 106.59 m^2.$

Thus, the minimum area of sheet metal required is approximately $106.59 m^2.$