Cambridge Additional Mathematics

DEMO

3

2

q

A q

B

a

b

q

B

A

2 m

3 m

0° 90°

• 
  
  
  
  • 41.1° μ
  
  
  
  

d

d__L~

q

20 cm

10 cm

20 cm

Applications of differential calculus (Chapter 14) 395

Step 5: The minimum material is used to make the container when x=20and y=

4000

202

=10.

Step 6: The most economical shape has a square base

20 cm£ 20 cm, and height 10 cm.

Example 20 Self Tutor

Two corridors meet at right angles and are 2 m and 3 m wide

respectively. μis the angle marked on the given figure. [AB]

is a thin metal tube which must be kept horizontal and cannot

be bent as it moves around the corner from one corridor to the

other.

a Show that the length AB is given by L=

3

cosμ

+

2

sinμ

.

b Show that

dL

dμ

=0when μ= tan¡^1

³

3

q

2

3

́

¼ 41 : 1 ±.

c FindLwhen μ= tan¡^1

³

3

q

2

3

́

and comment on the significance of this value.

a cosμ=

3

a

and

) a=

3

cosμ

and

) L=a+b=

3

cosμ

+

2

sinμ

sinμ=

2

b

b=

2

sinμ

b L= 3[cosμ]¡^1 + 2[sinμ]¡^1

)

dL

dμ

=¡3[cosμ]¡^2 (¡sinμ)¡2[sinμ]¡^2 cosμ

=

3 sinμ

cos^2 μ

¡

2 cosμ

sin^2 μ

=

3 sin^3 μ¡2 cos^3 μ

cos^2 μsin^2 μ

Thus

dL

dμ

=0 when 3 sin^3 μ= 2 cos^3 μ

) tan^3 μ=^23

) tanμ=^3

q

2

3

) μ= tan¡^1

μ

3

q

2

3

¶

¼ 41 : 1 ±

c Sign diagram of

dL

dμ

:

When μ=30±,

dL

dμ

¼¡ 4 : 93 < 0

When μ=60±,

dL

dμ

¼ 9 : 06 > 0

Thus, AB is minimised when μ¼ 41 : 1 ±. At this time L¼ 7 : 02 metres. Ignoring the width of

the rod, the greatest length of rod able to be horizontally carried around the corner is 7 : 02 m.

4037 Cambridge

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Y:\HAESE\CAM4037\CamAdd_14\395CamAdd_14.cdr Monday, 7 April 2014 12:19:28 PM BRIAN