Page 38 - Ranger SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation
πj 3
(b) I = 150πj – 2 Solution
dI 3πj 2 (a)
dj = 150π – 2
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dI h cm
When = 0,
dj x x cm 6 cm
cm
3πj 2
150π – 2 = 0 t cm
3πj 2
2 = 150π 2 2 cm
j = 150π × 3π h x
2
= 100 6 = 2
j = 10 cm h = 3x
d I Thus, the height of the cylinder
2
(c) = –3πj t = 6 – h
dj 2
When j = 10 cm, = 6 – 3x
d y The volume of the cylinder
2
= –3π(10)  0
dx 2 V = πx t
2
Hence, the volume of the container = πx (6 – 3x)
2
is maximum when j = 10 cm. = 6πx – 3πx 3
2
π(10)
I = 2 [300 – (10) ] (b) dV = 12πx – 9πx 2
2
dx
= 1 000π cm 3 For the volume to be maximum
dV
HOTS Example 1 dx = 0
The diagram below shows a cone with a 12πx – 9πx = 0
2
radius of 2 cm and a height of 6 cm on a 3πx(4 – 3x) = 0
horizontal table. A cylinder with a radius of 4 – 3x = 0
x cm lies inside the cone and touches the x = 4
surface of the cone and its axis of symmetry 3
2
coincides with the axis of symmetry of the d V = 12π – 18πx
cone. Given that the volume of the cylinder dx 2 4
is V cm . When x = ,
3
3
d V 4
2
dx 2 = 12π – 18π  
3
= –12π  0
6 cm Form 5
4
Hence, V is maximum when x = .
3
x cm 4 2 4 3
3
3
2 cm V = 6π   – 3π  
(a) Show that V = 6πx – 3πx . = 32 π cm 3
2
3
(b) When the value of x changes, find the 9
maximum volume of the cylinder in
terms of π.
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02 Ranger Mate Tambahan Tg5.indd 191 25/02/2022 9:23 AM