Page 65 - Hybrid PBD 2022 Tg 5 - Matematik Tambahan
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Matematik Tambahan Tingkatan 5 Bab 2 Pembezaan
1. A 8 m D
x m
F G
B E x m C
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Rajah menunjukkan sebuah dinding segi empat sama ABCD dengan Ungkapkan luas pintu
ukuran 8 m × 8 m. CEFG ialah pintu. Bahagian dinding akan dicatkan. Diberi dalam sebutan x. Untuk luas
d L
2
EC = GD = x m. KBAT Menganalisis minimum, dx 2 . 0.
The diagram shows a square wall ABCD with a dimension of 8 m × 8 m. CEFG is a door. The shaded Express the area of door in terms
2
region is to be painted. Given EC = GD = x m. of x. For minimum area, d L 2 . 0.
dx
(a) Tunjukkan bahawa dinding akan dicat mempunyai luas L = 64 – 8x + x 2.
Show that the wall to be painted has an area of L = 64 – 8x + x .
2
(b) Cari nilai x supaya luas dinding dicat adalah minimum.
Find the value of x so that the wall to be painted is minimum.
(c) Apakah harga perbelanjaan mengecat jika kos bagi satu meter persegi
ialah RM3.50.
What is the cost of painting the wall if it costs RM35 for one square metre.
(a) L = 8 – x(8 – x) (c) L = 64 – 8(4) + 4 = 48 m 2
2
2
= 64 – 8x + x 2 Kos/Cost = 48 × RM3.5 = RM168
(b) dL = –8 + 2x
dx
Untuk maksimum/minimum
dL = 0 ; x = 4 m
dx
2
d L = 2 . 0 ∴ minimum Sudut dicangkum ialah 90°,
1
dx 2 maka luas sektor = pj .
2
4
Untuk luas maksimum,
2. Rajah menunjukkan dua buah bulatan sepusat O. Jejari d L 2 , 0.
2
untuk bulatan besar ialah 8 cm. Perbezaan jejari di antara dx
bulatan besar dan kecil ialah 2x cm. The subtended angle is 90°, hence
1
2
The diagram shows two circles with the same centre O. The radius of the O 2x the area of sector = pj . For
4
2
big circle is 8 cm. The difference between the radius of the big circle and the 8 cm maximum area, d L 2 , 0 .
small circle is 2x cm. KBAT Mengaplikasi dx
(a) Tunjukkan bahawa luas rantau berlorek L ialah
L = p(8x – x ).
2
Show that the area of the shaded region L is L = p(8x – x ).
2
(b) Seterusnya, cari nilai x supaya luas maksimum tercapai dan apakah luas
maksimum itu?
Then, find the value of x so that the maximum area is obtained and what is the maximum area?
1
2
(a) L = p[8 – (8 – 2x) ] (b) dL = p[8 – 2x] = 0
2
4 dx x = 4 cm
1 d L
2
2
= p[32x – 4x ] = p[–2] , 0 ∴ maksimum / maximum
4 dx 2
= p[8x – x ] L = p[8(4) – 4 ]
2
2
= 16 p cm 2 Kuiz 2
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02 Hybrid PBD Mate Tamb Tg5.indd 44 09/11/2021 9:24 AM
