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

The cost of the base is: $C_{base} = 15x^2$

The cost of the sides (there are four sides) is: $C_{sides} = 8(4xh) = 32xh$

Thus, the total cost function $C$ becomes: $C = 15x^2 + 32x \left( \frac{60}{x^2} \right)$

This simplifies to: $C = 15x^2 + \frac{1920}{x}$

To find the minimum cost, we differentiate $C$ with respect to $x$ and set it to zero: $\frac{dC}{dx} = 30x - \frac{1920}{x^2} = 0$

Solving for $x$, we get: $30x^3 = 1920$ $x^3 = 64$ $x = 4$

Now substituting $x = 4$ back to find $h$: $h = \frac{60}{4^2} = \frac{60}{16} = 3.75$

Therefore, the height of the tank that minimizes the cost is $h = 3.75$ m.

To account for the thickness of the base and sides (2.5 cm or 0.025 m), you would adjust the dimensions:

1. The effective height would become $h - 0.025 = 3.725$ m.
2. The base length would reduce slightly, as it would be $x - 2 \times 0.025 = 3.95$ m.

This would result in recalculating the volume and consequently adjusting the cost function, but the overall effect would likely be minimal relative to the original dimensions.

The refinement would likely increase the overall cost, as the effective dimensions would yield less volume. This would result in a slight increase in costs, but due to the high volume, the impact on the cost minimization would not be significant.