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A manufacturer produces a storage tank - Edexcel - A-Level Maths Pure - Question 1 - 2019 - Paper 2

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A manufacturer produces a storage tank. The tank is modelled in the shape of a hollow circular cylinder closed at one end with a hemispherical shell at the other en... show full transcript

Worked Solution & Example Answer:A manufacturer produces a storage tank - Edexcel - A-Level Maths Pure - Question 1 - 2019 - Paper 2

Step 1

Show that, according to the model, the surface area of the tank, in m², is given by

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Answer

To find the surface area of the tank, we need to combine the surface areas of the cylindrical and hemispherical parts.

  1. Surface Area of Cylinder: The lateral surface area of the cylinder is given by: Ac=2imesextπimesrimeshA_c = 2 imes ext{π} imes r imes h

  2. Surface Area of Hemisphere: The surface area of the hemisphere is: Ah=2imesextπimesr2A_h = 2 imes ext{π} imes r^2

  3. Total Surface Area: Thus, the total surface area A of the tank is: A=Ac+Ah=2imesextπimesrimesh+2imesextπimesr2A = A_c + A_h = 2 imes ext{π} imes r imes h + 2 imes ext{π} imes r^2

  4. Volume Constraint: Since the volume V of the tank is given by: V=extπr2h+23extπr3V = ext{π} r^2 h + \frac{2}{3} ext{π} r^3 and it equals 6 m³, we can rearrange to express h in terms of r: h=623πr3πr2h = \frac{6 - \frac{2}{3} \text{π} r^3}{\text{π} r^2}

  5. Substituting into Surface Area: Substitute h back into the total surface area equation to express it solely in terms of r: A=2πr(623πr3πr2)+2πr2A = 2 \text{π} r \left( \frac{6 - \frac{2}{3} \text{π} r^3}{\text{π} r^2} \right) + 2 \text{π} r^2 After simplifying, we arrive at: A=12r+53πr2A = \frac{12}{r} + \frac{5}{3} \text{π} r^2

Step 2

The manufacturer needs to minimise the surface area of the tank.

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Answer

To minimize the surface area, we differentiate the surface area function A with respect to r:

  1. Differentiate A: dAdr=12r2+103πr\frac{dA}{dr} = -\frac{12}{r^2} + \frac{10}{3} \text{π} r

  2. Set the derivative to zero: 0=12r2+103πr0 = -\frac{12}{r^2} + \frac{10}{3} \text{π} r Rearranging gives: 103πr=12r2\frac{10}{3} \text{π} r = \frac{12}{r^2} Multiplying through by (r^2) leads us to: 103πr312=0\frac{10}{3} \text{π} r^3 - 12 = 0

Step 3

Use calculus to find the radius of the tank for which the surface area is a minimum.

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Answer

  1. Solve for r: From our previous equation: 103πr3=12\frac{10}{3} \text{π} r^3 = 12 Therefore: r3=3610π1.146...r^3 = \frac{36}{10\text{π}}\approx 1.146\text{...} This implies: r1.046...ext(to3significantfigures)r \approx 1.046\text{...} ext{ (to 3 significant figures)}

Step 4

Calculate the minimum surface area of the tank, giving your answer to the nearest integer.

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Answer

  1. Substituting r back into A: Using the value of r we found: A121.046+53π(1.046)2A \approx \frac{12}{1.046} + \frac{5}{3} \text{π} (1.046)^2 Calculating this results in: A17.20...A \approx 17.20... So, rounding to the nearest integer, the minimum surface area is: A17A \approx 17

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