User interface language: English | Español

Date May 2017 Marks available 5 Reference code 17M.1.AHL.TZ2.H_10
Level Additional Higher Level Paper Paper 1 Time zone Time zone 2
Command term Find and Justify Question number H_10 Adapted from N/A

Question

A window is made in the shape of a rectangle with a semicircle of radius r metres on top, as shown in the diagram. The perimeter of the window is a constant P metres.

M17/5/MATHL/HP1/ENG/TZ2/10

Find the area of the window in terms of P and r .

[4]
a.i.

Find the width of the window in terms of P when the area is a maximum, justifying that this is a maximum.

[5]
a.ii.

Show that in this case the height of the rectangle is equal to the radius of the semicircle.

[2]
b.

Markscheme

the width of the rectangle is 2 r and let the height of the rectangle be h

P = 2 r + 2 h + π r      (A1)

A = 2 r h + π r 2 2      (A1)

h = P 2 r π r 2

A = 2 r ( P 2 r π r 2 ) + π r 2 2 ( = P r 2 r 2 π r 2 2 )      M1A1

[4 marks]

a.i.

d A d r = P 4 r π r      A1

d A d r = 0      M1

r = P 4 + π      (A1)

hence the width is 2 P 4 + π      A1

d 2 A d r 2 = 4 π < 0      R1

hence maximum     AG

[5 marks]

a.ii.

EITHER

h = P 2 r π r 2

h = P 2 P 4 + π P π 4 + π 2      M1

h = 4 P + π P 2 P π P 2 ( 4 + π )     A1

h = P ( 4 + π ) = r      AG

OR

h = P 2 r π r 2

P = r ( 4 + π )      M1

h = r ( 4 + π ) 2 r π r 2      A1

h = 4 r + π r 2 r π r 2 = r      AG

[2 marks]

b.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.

Syllabus sections

Topic 5—Calculus » SL 5.6—Stationary points, local max and min
Show 140 related questions
Topic 5—Calculus » SL 5.7—Optimisation
Topic 5—Calculus

View options