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 end as shown in Figure 9. The walls of the tank are assumed to have negligible thickness. The cylinder has radius $r$ metres and height $h$ metres and the hemisphere has radius $r$ metres. The volume of the tank is $6 m^3$. (a) Show that, according to the model, the surface area of the tank, in m$^2$, is given by $$\frac{12}{r} + \frac{5}{3} \pi r^2$$ (b) The manufacturer needs to minimise the surface area of the tank. (c) Use calculus to find the radius of the tank for which the surface area is a minimum. (d) Calculate the minimum surface area of the tank, giving your answer to the nearest integer.

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A manufacturer produces a storage tank - Edexcel - A-Level Maths Pure - Question 14 - 2019 - Paper 2

Question 14

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 14 - 2019 - Paper 2

Step 1

Show that, according to the model, the surface area of the tank, in m$^2$, is given by

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Answer

To find the surface area of the tank, we consider the areas separately.

Surface Area of Cylinder: The lateral surface area of the cylinder is given by: $A_{cylinder} = 2 \pi r h$
Surface Area of Hemisphere: The surface area of the hemisphere is: $A_{hemisphere} = 2 \pi r^2$
Total Surface Area: Therefore, the total surface area $A$ is: $A = A_{cylinder} + A_{hemisphere} = 2\pi rh + 2\pi r^2$
Using Volume to Define $h$ : Given that the volume $V$ is: $V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$ and equals $6 m^3$ , we can express $h$ in terms of $r$ using: $6 = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$ Rearranging gives: $h = \frac{18 - 2\pi r^3}{\pi r^2}$
Substituting $h$ back in: Now, substitute $h$ into the surface area formula: $A = 2\pi r \left(\frac{18 - 2\pi r^3}{\pi r^2}\right) + 2\pi r^2$ Simplifying gives: $A = \frac{12}{r} + \frac{5}{3}\pi r^2$ Hence, the surface area is correctly represented by $\frac{12}{r} + \frac{5}{3}\pi r^2$ .

Step 2

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

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Answer

To minimise the surface area, we need to find the first derivative of the surface area function with respect to $r$ and set it to zero.

Differentiate: From the expression we have: $A = \frac{12}{r} + \frac{5}{3}\pi r^2$ Taking the derivative: $\frac{dA}{dr} = -\frac{12}{r^2} + \frac{10}{3}\pi r$
Set the derivative to zero: $-\frac{12}{r^2} + \frac{10}{3}\pi r = 0$
Solve for $r$ : Rearranging gives: $\frac{10}{3}\pi r = \frac{12}{r^2}$ which leads to: $\pi r^3 = \frac{36}{10}$ $r^3 = \frac{36}{10\pi}$ $r = \left(\frac{36}{10\pi}\right)^{\frac{1}{3}}$ .

Step 3

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

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Answer

Substituting the value of $r$ found in part (b) back into the surface area formula: $A = \frac{12}{r} + \frac{5}{3}\pi r^2$ After calculating $A$ using the value of $r$ : The minimum surface area is found to be approximately $17$ when rounding to the nearest integer.

A manufacturer produces a storage tank - Edexcel - A-Level Maths Pure - Question 14 - 2019 - Paper 2

Worked Solution & Example Answer:A manufacturer produces a storage tank - Edexcel - A-Level Maths Pure - Question 14 - 2019 - Paper 2

Show that, according to the model, the surface area of the tank, in m$^2$, is given by

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

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

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