Page 56 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan

Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan

(b) Anda diminta menentukan perjalanan paling murah Jalan raya/ Road 8 km Bandar/ Town
bagi kabel telefon yang menyambung sebuah sekolah S Kabel x Ladang kelapa sawit 2 km
ke bandar O seperti ditunjukkan dalam rajah. Cable Palm oil estate
You are asked to determine the cheapest route for a telephone cable that Sekolah/ School
connects a school S to the town O as shown in the diagram.
(i) Jika kos menyambung kabel di tepi jalan ialah RM5 000 sekilometer dan RM8 000 sekilometer
melalui ladang kelapa sawit, cari kos terendah.
If the cost to lay the cables by the road side is RM5 000 per km and RM8 000 per km across the palm oil estate, find the least
cost.
(ii) Apakah kos terendah jika harga kos kabel ialah RM7 000 sekilometer ditambah kepada pembinaan
tersebut.
What is the least cost if the cable cost is RM7 000 per km is added to cost to lay the cables.
(i) Kos/Cost C = (8 – x)5 000 + 8 000 2 + x
 2
2
d yPenerbitan Pelangi Sdn Bhd. All Rights Reserved.
dC 8 000(2x)
= –5 000 +

dx 2 4 + x 2
dC
Untuk minimum, = 0
dx
4 000(2x)
5 000 = 
4 + x
2
100 = 64x – 25x 2
2
100
x =  = 1.6 km
39
Kos terendah/The least cost C = (8 – 1.6)5 000 + 8 000 2 + 1.6
2
 2
= RM52 490
(ii) Panjang kabel/Length of cable = (8 - 1.6) + 2 + 1.6 = 8.96 km
 2
2
Jumlah kos/Total cost = 8.96 × RM7 000 + RM52 490 = RM115 210
23. Bagi setiap persamaan yang berhubung dengan x dan y berikut, jika kadar perubahan x ialah 0.04 unit s , cari
–1
kadar perubahan y pada ketika nilai x yang diberi. Tafsirkan jawapan anda. SP 2.4.6 TP4
For each of the equation relating x and y, if the rate of change of x is 0.04 unit s , find the rate of change of y at the given value of x. Interpret
–1
your answers.
(a) y = (2x + 3) , x = –1 x
3
1
y = x + 6 , x = (b) y = x – 2 , x = 3
2
2 d x
d x Diberi = 0.04 unit/s apabila d x
Diberi = 0.04 unit/s apabila Given dt when Diberi dt = 0.04 unit/s apabila
Given dt when x = –1 Given when
1 d y x = 3
x = = 3(2x + 3) (2) d y [(x – 2) – x] –2
2
2
d y = 2x dx dx = (x – 2) 2 = (x – 2) 2
dt Maka/Then dy = d y × d x d y d y d x
Maka/Then dt dx dt d x Maka/Then dt = dx × dt
2
d x
d y = × d x = 2x   = 6(2x + 3) dt –2 2  
d x
dt dx dt dt Apabila x = –1 dan d x = 0.04 = (x – 2) dt
1 dt
Apabila/When x = Apabila/When x = 3
2
d x Maka/Then Maka/Then
dan/and = 0.04 d y 2 d y –2
dt dt = 6(2(–1) + 3) (0.04) = (0.04)
Maka/Then = 0.24 unit/s dt (3 – 2) 2
1
d y = 2   (0.04) = 0.04 unit/s y bertambah dengan kadar = –0.08 unit/s
dt 2 0.24 unit/s y menyusut dengan kadar
y bertambah dengan kadar y increases at a rate of 0.24 unit/s y decreases at a rate of
y increases at a rate of 0.08 unit/s
0.04 unit/s

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