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Date	May 2018	Marks available	3	Reference code	18M.2.SL.TZ1.T_6
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Show that	Question number	T_6	Adapted from	N/A

Question

A manufacturer makes trash cans in the form of a cylinder with a hemispherical top. The trash can has a height of 70 cm. The base radius of both the cylinder and the hemispherical top is 20 cm.

A designer is asked to produce a new trash can.

The new trash can will also be in the form of a cylinder with a hemispherical top.

This trash can will have a height of H cm and a base radius of r cm.

There is a design constraint such that H + 2r = 110 cm.

The designer has to maximize the volume of the trash can.

Write down the height of the cylinder.

[1]

Find the total volume of the trash can.

[4]

Find the height of the cylinder, h , of the new trash can, in terms of r.

[2]

Show that the volume, V cm³ , of the new trash can is given by

$V = 110\pi {r^3}$ .

[3]

Using your graphic display calculator, find the value of r which maximizes the value of V.

[2]

The designer claims that the new trash can has a capacity that is at least 40% greater than the capacity of the original trash can.

State whether the designer’s claim is correct. Justify your answer.

[4]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

50 (cm) (A1)

[1 mark]

$\pi \times 50 \times {20^2} + \frac{1}{2} \times \frac{4}{3} \times \pi \times {20^3}$ (M1)(M1)(M1)

Note: Award (M1) for their correctly substituted volume of cylinder, (M1) for correctly substituted volume of sphere formula, (M1) for halving the substituted volume of sphere formula. Award at most (M1)(M1)(M0) if there is no addition of the volumes.

$= 79600\,\,\left( {{\text{c}}{{\text{m}}^3}} \right)\,\,\left( {79587.0 \ldots \left( {{\text{c}}{{\text{m}}^3}} \right)\,,\,\,\frac{{76000}}{3}\pi } \right)$ (A1)(ft) (G3)

Note: Follow through from part (a).

[4 marks]

h = H − r (or equivalent) OR H = 110 − 2r (M1)

Note: Award (M1) for writing h in terms of H and r or for writing H in terms of r.

(h =) 110 − 3r (A1) (G2)

[2 marks]

$\left( {V = } \right)\,\,\,\,\frac{2}{3}\pi {r^3} + \pi {r^2} \times \left( {110 - 3r} \right)$ (M1)(M1)(M1)

Note: Award (M1) for volume of hemisphere, (M1) for correct substitution of their h into the volume of a cylinder, (M1) for addition of two correctly substituted volumes leading to the given answer. Award at most (M1)(M1)(M0) for subsequent working that does not lead to the given answer. Award at most (M1)(M1)(M0) for substituting H = 110 − 2r as their h.

$V = 110\pi {r^2} - \frac{7}{3}\pi {r^3}$ (AG)

[3 marks]

(r =) 31.4 (cm) (31.4285… (cm)) (G2)

$\left( \pi \right)\left( {220r - 7{r^2}} \right) = 0$ (M1)

Note: Award (M1) for setting the correct derivative equal to zero.

(r =) 31.4 (cm) (31.4285… (cm)) (A1)

[2 marks]

$\left( {V = } \right)\,\,\,\,110\pi {\left( {31.4285 \ldots } \right)^3} - \frac{7}{3}\pi {\left( {31.4285 \ldots } \right)^3}$ (M1)

Note: Award (M1) for correct substitution of their 31.4285… into the given equation.

= 114000 (113781…) (A1)(ft)

Note: Follow through from part (e).

(increase in capacity =) $\frac{{113.781 \ldots - 79587.0 \ldots }}{{79587.0 \ldots }} \times 100 = 43.0\,\,\left( {\text{% }} \right)$ (R1)(ft)

Note: Award (R1)(ft) for finding the correct percentage increase from their two volumes.

1.4 × 79587.0… = 111421.81… (R1)(ft)

Note: Award (R1)(ft) for finding the capacity of a trash can 40% larger than the original.

Claim is correct (A1)(ft)

Note: Follow through from parts (b), (e) and within part (f). The final (R1)(A1)(ft) can be awarded for their correct reason and conclusion. Do not award (R0)(A1)(ft).

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.1—3d space, volume, angles, midpoints

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Topic 3—Geometry and trigonometry

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