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4047 4AM Differentiation (9) Math Academy®
©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any
retrieval system of any nature without prior permission.
Math Academy®
1
“If you make a mistake and do not correct it, this is called a mistake.”
― Confucius
Application of Differentiation – Maximum and Minimum Problems
[A] With Magic Number
The diagram shows a box in the shape of a cuboid with a square cross-section of side cm.
The volume of the box is 3500 cm . Four pieces of tape are fastened round the box as
shown. The pieces of tape are parallel to the edges of the box.
(i) Given that the total length of the four pieces of tape is cm, show that
. [3]
(ii) Given that can vary, find the stationary value of and determine the nature of this
stationary value. [5]
Solution:
(i) Use the magic number 3500 to form an equation (ii)
Let the length of the cuboid be cm.
Total length
(shown)
Ans: (ii) L = 210, minimum
x
3
L
2
7000
14
x
xL +=
x L
y
2
3500x y =
2
3500
y
x
=
L 14 2x y= +
2
3500
14 2x
x
æ ö
= + ç ÷
è ø
2
7000
14x
x
= +
3D
0
dL
dx
=
Example 1:
Observe and
make the
subject.
y
4047 4AM Differentiation (9) Math Academy®
©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any
retrieval system of any nature without prior permission.
Math Academy®
2
[B] Without Magic Number (may need to draw extra lines to form )
Example 2: The diagram shows a right circular cone with radius cm and height cm, where
and can vary. A sphere with radius 9 cm is to be placed inside the cone, such
that it will touch the top surface of the cone.
(i) Show that the volume of the cone, cm , can be expressed as .
[3]
(ii) Hence, find the minimum volume of the cone. [5]
Solution:
Ans: (ii) , min Vol = 6110 cm (3sf)
r h
r h
V 3
2
27
18
h
V
h
p
=
-
36h = 3
2D
Join from centre
of circle to tangent
Similar triangles /
Toa Cah Soh /
Pythagoras Theorem
0
dV
dh
=
cm
9 cm
cm
Right Angled Triangle
4047 4AM Differentiation (9) Math Academy®
©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any
retrieval system of any nature without prior permission.
Math Academy®
3
Our greatest weakness lies in giving up.
The most certain way to succeed is always to try just one more time.
Thomas A. Edison
Worksheet 7: Maximum and Minimum Problems
Qns compiled from exam papers
1 A piece of wire, 48 cm in length, is used to form the framework of a regular prism
as shown in the diagram below. The cross-section of the prism is a right-angled
triangle with two perpendicular sides of length cm and cm.
The length of the prism is cm.
(a) Express in terms of . [2]
(b) Show that the volume of the prism
is given by . [2]
(c) Find the value of for which has a stationary value. [3]
(d) Find this value of and determine whether it is a maximum or a minimum.
[3]
Solution:
x
1
1
3
x
y
y x
216
(6 )
9
V x x= -
x V
V
x cm
1 x cm
y cm
Magic no.
(a) By pythagoras theorem
Length = 48 cm
(b) Volume
2
2 21
1
3
x x w
æ ö
+ =ç ÷
è ø
2 225
9
x w=
5
3
x
w =
1 5
2 2 1 2 3 48
3 3
x
x x y
æ ö æ ö
+ + + =ç ÷ ç ÷
è ø è ø
8 3 48x y+ =
8
16
3
y x= -
1 1
( ) 1 ( )
2 3
x x y
æ ö
= ç ÷
è ø
22 8
16
3 3
x x
æ ö
= -ç ÷
è ø
216
(6 )
9
x x= -
(c)
or
(rejected)
(d) cm
When ,
Volume is a maximum at
2 332 16
3 9
V x x= -
264 16
3 3
dV
x x
dx
= -
64 16
0
3 3
x x
æ ö
- =ç ÷
è ø
0x = 4x =
216 8
(4) (6 4) 56
9 9
V = - = 3
2
2
64 32
3 3
d V
x
dx
= -
4x =
2
2
64
0
3
d V
dx
= - <
 4x =
4047 4AM Differentiation (9) Math Academy®
©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any
retrieval system of any nature without prior permission.
Math Academy®
4
3 The diagram shows a cylinder of height cm and base radius cm inscribed in a sphere
of radius 35 cm.
(i) Show that the height of the cylinder, cm is given by . [2]
(ii) Given that can vary, find the maximum volume of the cylinder. [4]
Ans: (ii) 104 000 cm
Solution:
h r
h 2
2 1225h r= -
r
3
positive 0 negative
Slope
r 28.5 28.577 28.6
dT
dx
(i) By Pythagoras’ Theorem,
(shown)
(ii)
(rejected)
Volume = 104 000 cm (3 sf)
Volume is a maximum when .
2
2 2
35
2
h
r
æ ö
+ =ç ÷
è ø
2
2
1225
4
h
r= -
2 2
4(1225 )h r= -
2
2 1225h r= -
2 2
(2 1225 )V r rp= -
( )
1
2 2 21225r rp= -
( ) ( )
1 1
2 2 22 2
1
2 1225 ( 2 ) 1225 (4 )
2
dV
r r r r r
dr
p p
-æ ö
= - - + -ç ÷
è ø
( ) ( )
1 1
3 2 22 22 1225 1225 (4 )r r r rp p
-
= - - + -
( ) ( )
1
2 3 221225 2 (1225 )(4 )r r r rp p
-
= - - + -
( ) ( )
1
2 2 222 1225 2450 2r r r rp
-
= - - + -
( ) ( )
1
2 222 1225 3 2450r r rp
-
= - - +
1
2
2
2
2 ( 3 2450)
0
(1225 )
r r
r
p - +
=
-
0r = 2
3 2450 0r- + =
28.577r =
3
 28.577r =
Keyword Max/min qn®
4047 4AM Differentiation (9) Math Academy®
©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any
retrieval system of any nature without prior permission.
Math Academy®
5
5
The diagram shows a cuboid of height units inside a right pyramid of height
8 units and with square base of side 4 units. The base of the cuboid sits on the square base
of the pyramid. The points and are corners of the cuboid and lie on the
edges and , respectively, of the pyramid The pyramids
and are similar.
(i) Find an expression for in terms of and hence show that the volume
of the cuboid is given by units . [4]
(ii) Given that can vary, find the value of for which is a maximum.
[4]
Solutions:
h OPQRS
PQRS A, B, C D
OP, OQ, OR OS OPQRS. OPQRS
OABCD
AD h V
V =
h3
4
− 4h2
+16h 3
h h V
Similar Triangles
(i) By similar triangles,
(ii) Volume of cuboid =
8
4 8
AD h-
=
8
2
h
AD
-
=
2
8
2
h
h
-æ ö
= ç ÷
è ø
2
(64 16 )
4
h h h- +
=
3
2
4 16
4
h
h h= - +
(ii)
or
(rejected)
When ,
Volume is a maximum when .
2
3
8 16 0
4
dV h
h
dh
= - + =
( 8)(3 8) 0h h- - =
8h =
8
3
h =
2
2
3
8
2
d V h
dh
= -
8
3
h =
2
2
4 0
d V
dh
= - <

8
3
h =
4047 4AM Differentiation (9) Math Academy®
©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any
retrieval system of any nature without prior permission.
Math Academy®
6
7 A lifeguard at a beach resort is stationed at point along the coastline, as shown in the
diagram below. When he detects a swimmer who needs help at a point , he would run
along the coastline over a distance of m to a point , and then swim in a straight line,
, towards the swimmer. The lifeguard runs at a speed of 4 m/s and swims at a speed
of 2 m/s.
A swimmer in distress is detected at a position that is 40 m away from the coastline,
and the foot of the perpendicular from the swimmer to the coastline is at a distance of
60 m away from the lifeguard.
(i) Show that the time taken by the lifeguard to swim from to is
seconds. [2]
(ii) Find, in terms of , the total time taken by the lifeguard to reach the
swimmer. [1]
(iii) Obtained an expression for . [2]
(iv) Find the value of such that the lifeguard would be able to reach the
swimmer in the shortest possible time. [4]
Ans: (ii) (iii) 36.9
G
S
x H
HS
H S
2
1600 (60 )
2
x+ -
x T
dT
dx
x
2
60 1
42 1600 (60 )
dT x
dx x
-
= +
+ -
Keyword Max/min qn®
4047 4AM Differentiation (9) Math Academy®
©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any
retrieval system of any nature without prior permission.
Math Academy®
7
9 At 8 a.m, Ship is 100 km due North of Ship .
Ship is sailing due South at 20 km/h.
Ship is sailing at a bearing of 120 at 10 km/h.
(i) Show that the distance between the two ships after hours is given by
. [3]
(ii) At what time is Ship closest to Ship . [3]
Solution: (i) Assume that the distance moved by Ship is less than 100 km.
Let’s draw the diagram out.
By Cosine Rule,
(shown)
(ii) For Ship closest to Ship ,
Ship is closest to Ship after 5 hours.
A B
A
B °
t
2
10 3 30 100AB t t= - +
A B
A
2 2 2
(100 20 ) (10 ) 2(100 20 )(10 )cos120AB t t t t= - + - - °
2 2 2 2
10000 4000 400 100 (2000 400 )( 0.5)AB t t t t t= - + + - - -
2 2
300 3000 10000AB t t= - +
2 2
100(3 30 100)AB t t= - +
2
10 3 30 100AB t t= - +
A B 0
dAB
dt
=
1
2 2
1
10 (3 30 100) (6 30) 0
2
t t t
-æ ö
- + - =ç ÷
è ø
6 30 0t - =
5t =
A B
Keyword Max/min qn®
100 20t-
10t

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Sec 4 A Maths Notes Maxima Minima

  • 1. 4047 4AM Differentiation (9) Math Academy® ©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. Math Academy® 1 “If you make a mistake and do not correct it, this is called a mistake.” ― Confucius Application of Differentiation – Maximum and Minimum Problems [A] With Magic Number The diagram shows a box in the shape of a cuboid with a square cross-section of side cm. The volume of the box is 3500 cm . Four pieces of tape are fastened round the box as shown. The pieces of tape are parallel to the edges of the box. (i) Given that the total length of the four pieces of tape is cm, show that . [3] (ii) Given that can vary, find the stationary value of and determine the nature of this stationary value. [5] Solution: (i) Use the magic number 3500 to form an equation (ii) Let the length of the cuboid be cm. Total length (shown) Ans: (ii) L = 210, minimum x 3 L 2 7000 14 x xL += x L y 2 3500x y = 2 3500 y x = L 14 2x y= + 2 3500 14 2x x æ ö = + ç ÷ è ø 2 7000 14x x = + 3D 0 dL dx = Example 1: Observe and make the subject. y
  • 2. 4047 4AM Differentiation (9) Math Academy® ©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. Math Academy® 2 [B] Without Magic Number (may need to draw extra lines to form ) Example 2: The diagram shows a right circular cone with radius cm and height cm, where and can vary. A sphere with radius 9 cm is to be placed inside the cone, such that it will touch the top surface of the cone. (i) Show that the volume of the cone, cm , can be expressed as . [3] (ii) Hence, find the minimum volume of the cone. [5] Solution: Ans: (ii) , min Vol = 6110 cm (3sf) r h r h V 3 2 27 18 h V h p = - 36h = 3 2D Join from centre of circle to tangent Similar triangles / Toa Cah Soh / Pythagoras Theorem 0 dV dh = cm 9 cm cm Right Angled Triangle
  • 3. 4047 4AM Differentiation (9) Math Academy® ©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. Math Academy® 3 Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time. Thomas A. Edison Worksheet 7: Maximum and Minimum Problems Qns compiled from exam papers 1 A piece of wire, 48 cm in length, is used to form the framework of a regular prism as shown in the diagram below. The cross-section of the prism is a right-angled triangle with two perpendicular sides of length cm and cm. The length of the prism is cm. (a) Express in terms of . [2] (b) Show that the volume of the prism is given by . [2] (c) Find the value of for which has a stationary value. [3] (d) Find this value of and determine whether it is a maximum or a minimum. [3] Solution: x 1 1 3 x y y x 216 (6 ) 9 V x x= - x V V x cm 1 x cm y cm Magic no. (a) By pythagoras theorem Length = 48 cm (b) Volume 2 2 21 1 3 x x w æ ö + =ç ÷ è ø 2 225 9 x w= 5 3 x w = 1 5 2 2 1 2 3 48 3 3 x x x y æ ö æ ö + + + =ç ÷ ç ÷ è ø è ø 8 3 48x y+ = 8 16 3 y x= - 1 1 ( ) 1 ( ) 2 3 x x y æ ö = ç ÷ è ø 22 8 16 3 3 x x æ ö = -ç ÷ è ø 216 (6 ) 9 x x= - (c) or (rejected) (d) cm When , Volume is a maximum at 2 332 16 3 9 V x x= - 264 16 3 3 dV x x dx = - 64 16 0 3 3 x x æ ö - =ç ÷ è ø 0x = 4x = 216 8 (4) (6 4) 56 9 9 V = - = 3 2 2 64 32 3 3 d V x dx = - 4x = 2 2 64 0 3 d V dx = - < 4x =
  • 4. 4047 4AM Differentiation (9) Math Academy® ©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. Math Academy® 4 3 The diagram shows a cylinder of height cm and base radius cm inscribed in a sphere of radius 35 cm. (i) Show that the height of the cylinder, cm is given by . [2] (ii) Given that can vary, find the maximum volume of the cylinder. [4] Ans: (ii) 104 000 cm Solution: h r h 2 2 1225h r= - r 3 positive 0 negative Slope r 28.5 28.577 28.6 dT dx (i) By Pythagoras’ Theorem, (shown) (ii) (rejected) Volume = 104 000 cm (3 sf) Volume is a maximum when . 2 2 2 35 2 h r æ ö + =ç ÷ è ø 2 2 1225 4 h r= - 2 2 4(1225 )h r= - 2 2 1225h r= - 2 2 (2 1225 )V r rp= - ( ) 1 2 2 21225r rp= - ( ) ( ) 1 1 2 2 22 2 1 2 1225 ( 2 ) 1225 (4 ) 2 dV r r r r r dr p p -æ ö = - - + -ç ÷ è ø ( ) ( ) 1 1 3 2 22 22 1225 1225 (4 )r r r rp p - = - - + - ( ) ( ) 1 2 3 221225 2 (1225 )(4 )r r r rp p - = - - + - ( ) ( ) 1 2 2 222 1225 2450 2r r r rp - = - - + - ( ) ( ) 1 2 222 1225 3 2450r r rp - = - - + 1 2 2 2 2 ( 3 2450) 0 (1225 ) r r r p - + = - 0r = 2 3 2450 0r- + = 28.577r = 3 28.577r = Keyword Max/min qn®
  • 5. 4047 4AM Differentiation (9) Math Academy® ©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. Math Academy® 5 5 The diagram shows a cuboid of height units inside a right pyramid of height 8 units and with square base of side 4 units. The base of the cuboid sits on the square base of the pyramid. The points and are corners of the cuboid and lie on the edges and , respectively, of the pyramid The pyramids and are similar. (i) Find an expression for in terms of and hence show that the volume of the cuboid is given by units . [4] (ii) Given that can vary, find the value of for which is a maximum. [4] Solutions: h OPQRS PQRS A, B, C D OP, OQ, OR OS OPQRS. OPQRS OABCD AD h V V = h3 4 − 4h2 +16h 3 h h V Similar Triangles (i) By similar triangles, (ii) Volume of cuboid = 8 4 8 AD h- = 8 2 h AD - = 2 8 2 h h -æ ö = ç ÷ è ø 2 (64 16 ) 4 h h h- + = 3 2 4 16 4 h h h= - + (ii) or (rejected) When , Volume is a maximum when . 2 3 8 16 0 4 dV h h dh = - + = ( 8)(3 8) 0h h- - = 8h = 8 3 h = 2 2 3 8 2 d V h dh = - 8 3 h = 2 2 4 0 d V dh = - < 8 3 h =
  • 6. 4047 4AM Differentiation (9) Math Academy® ©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. Math Academy® 6 7 A lifeguard at a beach resort is stationed at point along the coastline, as shown in the diagram below. When he detects a swimmer who needs help at a point , he would run along the coastline over a distance of m to a point , and then swim in a straight line, , towards the swimmer. The lifeguard runs at a speed of 4 m/s and swims at a speed of 2 m/s. A swimmer in distress is detected at a position that is 40 m away from the coastline, and the foot of the perpendicular from the swimmer to the coastline is at a distance of 60 m away from the lifeguard. (i) Show that the time taken by the lifeguard to swim from to is seconds. [2] (ii) Find, in terms of , the total time taken by the lifeguard to reach the swimmer. [1] (iii) Obtained an expression for . [2] (iv) Find the value of such that the lifeguard would be able to reach the swimmer in the shortest possible time. [4] Ans: (ii) (iii) 36.9 G S x H HS H S 2 1600 (60 ) 2 x+ - x T dT dx x 2 60 1 42 1600 (60 ) dT x dx x - = + + - Keyword Max/min qn®
  • 7. 4047 4AM Differentiation (9) Math Academy® ©All rights reserved. No part of this document may be reproduced or transmitted in any form or by any means, or stored in any retrieval system of any nature without prior permission. Math Academy® 7 9 At 8 a.m, Ship is 100 km due North of Ship . Ship is sailing due South at 20 km/h. Ship is sailing at a bearing of 120 at 10 km/h. (i) Show that the distance between the two ships after hours is given by . [3] (ii) At what time is Ship closest to Ship . [3] Solution: (i) Assume that the distance moved by Ship is less than 100 km. Let’s draw the diagram out. By Cosine Rule, (shown) (ii) For Ship closest to Ship , Ship is closest to Ship after 5 hours. A B A B ° t 2 10 3 30 100AB t t= - + A B A 2 2 2 (100 20 ) (10 ) 2(100 20 )(10 )cos120AB t t t t= - + - - ° 2 2 2 2 10000 4000 400 100 (2000 400 )( 0.5)AB t t t t t= - + + - - - 2 2 300 3000 10000AB t t= - + 2 2 100(3 30 100)AB t t= - + 2 10 3 30 100AB t t= - + A B 0 dAB dt = 1 2 2 1 10 (3 30 100) (6 30) 0 2 t t t -æ ö - + - =ç ÷ è ø 6 30 0t - = 5t = A B Keyword Max/min qn® 100 20t- 10t