Sec 4 A Maths Notes Maxima Minima

4047 4AM Differentiation (9) Math Academy®
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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

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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

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 =

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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®

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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 =

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
-
= +
+ -

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
100 20t-
10t

Sec 4 A Maths Notes Maxima Minima

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Sec 4 A Maths Notes Maxima Minima