An open-topped fish tank is to be made for an aquarium - AQA - A-Level Maths Mechanics - Question 14 - 2017 - Paper 1

Differentiating the total cost with respect to $x$ to find the minimum: $\frac{dC}{dx} = 30x - \frac{1920}{x^2}$ Set this equal to zero to find critical points: $30x - \frac{1920}{x^2} = 0$

Solving for $x$: $30x^3 = 1920$ $x^3 = 64 \Rightarrow x = 4$ Now, substituting back to find $h$: $h = \frac{60}{4^2} = \frac{60}{16} = 3.75$

Thus, the dimensions for the tank are $x = 4$ m (base length) and $h = 3.75$ m (height).

To refine the modelling considering the thickness of the sides and base:

• The width of the side and base should be adjusted to account for 2.5 cm thickness, which is equivalent to 0.025 m. This means that the effective lengths would be $x - 0.05$ for the sides, and $h - 0.025$ for the height.

The refinement is likely to affect the overall volume calculation, potentially leading to a slight change in the minimum cost values found in part (a). The effective dimensions would yield a slightly greater volume requirement, which could result in marginally higher costs.