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Date May 2018 Marks available 9 Reference code 18M.1.SL.TZ2.S_9
Level Standard Level Paper Paper 1 Time zone Time zone 2
Command term Find Question number S_9 Adapted from N/A

Question

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20 π cm3.

The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.

Express h in terms of r.

[2]
a.

Show that  C = 20 π r 2 + 320 π r .

[4]
b.

Given that there is a minimum value for C, find this minimum value in terms of π .

[9]
c.

Markscheme

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

correct equation for volume      (A1)
eg   π r 2 h = 20 π

h = 20 r 2      A1 N2

[2 marks]

 

a.

attempt to find formula for cost of parts      (M1)
eg  10 × two circles, 8 × curved side

correct expression for cost of two circles in terms of r (seen anywhere)      A1
eg   2 π r 2 × 10

correct expression for cost of curved side (seen anywhere)      (A1)
eg   2 π r × h × 8

correct expression for cost of curved side in terms of     A1
eg   8 × 2 π r × 20 r 2 , 320 π r 2

C = 20 π r 2 + 320 π r       AG N0

[4 marks]

b.

recognize C = 0  at minimum       (R1)
eg   C = 0 , d C d r = 0

correct differentiation (may be seen in equation)

C = 40 π r 320 π r 2        A1A1

correct equation      A1
eg   40 π r 320 π r 2 = 0 , 40 π r 320 π r 2

correct working     (A1)
eg   40 r 3 = 320 , r 3 = 8

r = 2 (m)     A1

attempt to substitute their value of r into C
eg   20 π × 4 + 320 × π 2      (M1)

correct working
eg   80 π + 160 π        (A1)

240 π  (cents)      A1 N3

Note: Do not accept 753.6, 753.98 or 754, even if 240 π is seen.

[9 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
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