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

Substituting equation (2) into the cost equation:

$C = 15x^2 + 32x \left( \frac{60}{x^2} \right) = 15x^2 + \frac{1920}{x} \quad (3)$

Next, we differentiate equation (3) with respect to $x$:

$\frac{dC}{dx} = 30x - \frac{1920}{x^2}$

Setting this to zero to find the critical points:

$30x - \frac{1920}{x^2} = 0 \quad (4)$

Rearranging gives:

$30x^3 = 1920 \iff x^3 = \frac{1920}{30} = 64 \iff x = 4$

Now substituting $x = 4$ back into equation (2) to find $h$:

$h = \frac{60}{4^2} = \frac{60}{16} = 3.75$

To ensure this is a minimum, we calculate the second derivative:

$\frac{d^2C}{dx^2} = 30 + \frac{3840}{x^3}$

Evaluating at $x = 4$:

$\frac{d^2C}{dx^2} = 30 + \frac{3840}{64} = 30 + 60 = 90 > 0$

Since the second derivative is positive, the cost function has a local minimum at $x = 4$ and hence the dimensions that minimize the cost are:

• Side length: $4$ m
• Height: $3.75$ m.

In reality, the thickness of the sides and base of the tank is 2.5 cm (0.025 m). This means that the effective dimensions of the tank must account for this thickness. The new side length would be:

$x_{effective} = x - 2(0.025) = x - 0.05$

Hence, I would revise equations (1) and (3) to reflect these adjusted dimensions.

This refinement would likely increase the overall cost as the dimensions are reduced by 0.05 m. Consequently, both the volume and surface area would be altered, potentially changing the values for $h$ and likely leading to an increase in the minimum cost found in part (a).