Page 64 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan
P. 64
Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
Jawapan / Answer : 2. 16 cm
y
x + y = 5 z = xy 2 y
2
2
= y (5 – y) = 5y – y 3
dz = 10y – 3y 2
dy = y[10 – 3y] = 0
y = 0 y = 10
3
x = 5 x = 5 Sekeping logam yang berbentuk segi empat
3 sama dengan bersempadan 16 cm, dilipat untuk
2
d z = –6y membentuk sebuah kotak terbuka setelah
dy 2 dikeluarkan 4 segi empat sama yang bersempadan
10
2
,
Apabila/When y = 10 d z = –6 = –20 , 0 y cm daripada setiap pepenjurunya.
3 dy 2 3 A square piece of metal has a side of 16 cm, is to be folded to
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5 10
∴ , ialah maksimum make an open box by cutting off 4 squares each with sides y cm
from each of the tips of the square.
3 3
(a) Apakah nilai y yang menjadikan kotak itu
mempunyai isi padu maksimum?
What is the value of y which will make the volume
maximum?
Kertas 2 [5 markah / 5 marks]
(b) Seterusnya, cari nilai isi padu maksimum itu.
Then, find the maximum volume.
1. Diberi bahawa persamaan suatu lengkung ialah [5 markah / 5 marks]
3
y = .
x 3
Given that the equation of a curve is y = x 3 3 . Jawapan/Answer:
2
(a) Cari nilai dy apabila x = 3. (a) V = (16 – 2y) y
dx dV
Find the value of dy when x = 3. = 2(16 – 2y)(–2)y + (16 – 2y) 2
dx dy = (16 – 2y)[–4y + 16– 2y]
[3 markah / 3 marks]
(b) Seterusnya, anggarkan nilai bagi 3 . = (16 – 2y)(16 – 6y)
3 (2.98) 3 dV
Then, estimate the value of . Apabila maksimum / minimum = 0
(2.98) 3 dy
[4 markah / 4 marks] When maximum / minimum
y = 8 atau 8
Jawapan/Answer: 3
(a) y = x 3 3 d V
2
dy = – 9 (b) dy 2 = (16 – 2y)(–6) + (16 – 6y)(–2)
dx x 4 = –2[48 – 6y + 16 – 2y]
= –2[64 – 8y]
Apabila/When x = 3 , dy = –9 = –1
2
dx 3 4 9 Apabila/When y = 8 , d V = 0
dy dy 2
(b) δy = .δx
8 d V
8
dx y = , 2 = –2 64 – 8
1
= – [–0.02] 3 dy 2 3
9 , 0 (Maksimum)
1 8
= = ∴ y = cm.
900 450 3
2 8
8
3 1 V = 16 – 2 = 303 11 cm 2
∴ y + δy = + 3 3 27
27 450
1
= + 1 = 17 = 0.113
9 450 150
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02 Hybrid PBD Mate Tamb Tg5.indd 43 09/11/2021 9:24 AM

