An open-topped fish tank is to be made for an aquarium. It will have a square horizontal base, rectangular vertical sides and a volume of 60 m³. The materials cost: • £15 per m² for the base • £8 per m² for the sides 14 (a) Modelling the sides and base of the fish tank as laminae, use calculus to find the height of the tank for which the overall cost of the materials has its minimum value. Fully justify your answer. 14 (b) (i) In reality, the thickness of the base and sides of the tank is 2.5 cm Briefly explain how you would refine your modelling to take account of the thickness of the sides and base of the tank of the tank. 14 (b) (ii) How would your refinement affect your answer to part (a)?

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

Question 14

An open-topped fish tank is to be made for an aquarium. It will have a square horizontal base, rectangular vertical sides and a volume of 60 m³. The materials cost... show full transcript

Worked Solution & Example Answer:An open-topped fish tank is to be made for an aquarium - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 1

Step 1

Modelling the cost with an expression

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Answer

Let the length of the sides of the base be denoted as $x$ . The height of the tank is $h$ . Given that the volume of the tank is 60 m³, the volume can be expressed as:

$V = x^2 \cdot h = 60 \quad (1)$

From equation (1), we can isolate $h$ :

$h = \frac{60}{x^2} \quad (2)$

Next, we need to determine the cost $C$ . The cost of the base is:

$C_{base} = 15 \cdot x^2$

The surface area of the vertical sides is given by the perimeter of the base times the height:

$C_{sides} = 8 \cdot (4xh) = 32xh$

Combining these, the total cost is:

$C = C_{base} + C_{sides} = 15x^2 + 32xh$

Step 2

Eliminating the variable

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Answer

Substituting equation (2) into the cost equation:

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

Step 3

Finding the minimum cost

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Answer

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 $h$ :

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

Step 4

Justifying the minimum cost

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Answer

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.

Step 5

Refining the modelling due to thickness

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Answer

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.

Step 6

Effect of the refinement on part (a)

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Answer

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).

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

Worked Solution & Example Answer:An open-topped fish tank is to be made for an aquarium - AQA - A-Level Maths Pure - Question 14 - 2017 - Paper 1

Modelling the cost with an expression

Eliminating the variable

Finding the minimum cost

Justifying the minimum cost

Refining the modelling due to thickness

Effect of the refinement on part (a)

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