Page 46 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

18 Solution
A wire with length 84 cm is bent to form a rectangle.
Find the maximum area of the rectangle that can be height = h
formed. width = x
Solution length = 2x
Area of rectangle = A From the analysis of the problem, the production
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Let A = length × width Width = y cost depends on the total surface area of the box
= x × y produced.
= xy Length = x
Therefore, A = total surface area
Length of wire = perimeter of the rectangle = 2[(2x)(x)] + 2(xh) + 2[(2x)(h)]
x + x + y + y = 84
2x + 2y = 84 h
2x = 84 – 2y x x h
x = 42 – y 2x 2x
2
Substitute into A, \ A = 4x + 2xh + 4xh
A = (42 – y)(y) Volume of the box given = 72 cm
3
= 42y – y Length × width × height = 72
2
dA = 42 – 2y 2x × x × h = 72
dy 72
At the turning point, h = 2x 2
0 = 42 – 2y = 36
2y = 42 x 2
y = 21 Substitute h = 36 into A,
x 2
2
36
36
Also, d A = –2 , 0 A = 4x + 2x 1 2 + 4x 1 2
2
dy 2 x 2 x 2
2
y = 21 will caused the value of A to be maximum. = 4x + 72 + 144
x
x
2
SPM Tips = 4x + 216
x
dA –2
A can also be expressed in terms of x to determine \ dx = 8x – 216x
the value of x.
At the turning point, 8x – 216 = 0
2
A max = 42(21) – (21) x 8x = 216
2
= 441 cm x 2
2
3
8x = 216
\ The maximum area of the rectangle that can be x = 27
3
formed is 441 cm . x = 3 Form 5
2
2
Try Questions 9 – 12 in ‘Try This! 2.4’ Also, d L = 8 + 432x
–3
dx 2
d L . 0
2
When x = 3,
Example of HOTS dx 2
HOTS Question

x = 3 will cause the value of A to be minimum.
A factory plans to produce a closed cuboid-shape 216
box such that the volume of the box produced is A = 4(3) + = 108 cm 2
2
min
72 cm . Given that the length of the box is twice its 3
3
width. The production cost of the box is estimated The minimum production cost of a box produced is
to be RM0.15 per cm . Find the minimum production = 108 × 0.15
2
cost that can be obtained in the production of a = RM16.20
box.
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