JABATAN PENDIDIKAN SELANGORKERJA PROJEK MATEMATIK TAMBAHAN 2015

RANGKA PENYELESAIAN TUGASANPART 1

(a) Accept any relevant answer(b) Accept any relevant answer

PART 2

(a)

Length of fence = 4x + 2y = 200 .................

Area , A = xy ........................

From

2x + y = 100

y = 100 – 2x ........

Substitute into

A = x( 100 – 2x)

= 100x – 2x2

dAdx

= 100 − 4 x

For maximum , dAdx

=0 , 100−4 x=0

x= 25

y = 100 – 2 (25) = 50

d 2 Adx2

=− 4

Therefore, maximum area = 25 X 50

= 1250m2

y

xx

y

xx

1

2

2

1

3

3

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b ) V = I × W × hV (h ) = (30−2h ) (30−2h )h

= 4 h3− 120h2 + 900 hV ' (h ) = 12h2−240h+90012h2−240h+900=0

h = 5 or 15

V(5)=2000

V(15)=0 Largest possible volume = 2000 cm3

PART 3

(a)(i)

(ii) 3.30 pm @ 1530

(iii) 3600 people

(iv)−1800 cos ( π t

6)+1800=2570

cos ( π6

t )=−0 . 4278

α =64 .67 °

π6

t =115. 33 ° , 244 .67 °

t = 3.844 , 8.156

1 2 3 4 5 6 7 8 9 10 11 12 13

3500

3000

2500

2000

1500

1000

500

-500

f x = -1800cosx

6 +1800

t

P( t)

FURTHER EXPLORATION

(a) Accept any relevant answer

(b) (i) (a) x : number of model x cabinets purchased

y : number of model y cabinets purchased

x≥ 0 , y ≥0

Cost : 100 x+200 y≤ 1400∨ y≤−12

x+7

Space : 0.6 x+0.8 y ≤7.2∨ y≤−34

x+9

No. of cabinets: y ≤32

x

Volume: v=0.8 x+1.2 y

(b) Refer to graph

(ii) Method (a):

Using the optimum linear line v=0.8 x+1.2 y . Maximum point (8,3)

Maximum volume =0.8(8)+1.2(3) = 10 cubic meters Method (b): Using the corner points

Points Volume (m3)(4,5) 0.8(4) + 1.2(5)=9.2(8,3) 0.8(8) + 1.2(3)=10.0

Maximum volume=10 cubic meters (iii)

Combination (x,y) Volume (m3) Cost (RM)(4,1) 4.4 600.00(4,2) 5.6 800.00(4,3) 6.8 1000.00(4,4) 8 1200.00(4,5) 9.2 1400.00(5,1) 5.2 700.00(5,2) 6.4 900.00(5,3) 7.6 1100.00(5,4) 8.8 1300.00(6,1) 6 800.00(6,2) 7.2 1000.00(6,3) 8.4 1200.00(6,4) 9.6 1400.00(7,1) 6.8 900.00(7,2) 8 1100.00(7,3) 9.2 1300.00(8,1) 7.6 1000.00

(8,2) 8.8 1200.00(8,3) 10 1400.00(9,1) 8.4 1100.00(9,2) 9.6 1300.00

(iv) Accept any relevant answer.

0 2 4 6 8 10 12 14 x

Graph y against x y

1

2

3

4

5

6

7

8

9

10

y=32

x

y=34

x+9

y=22

x+7

(8,3)

(4,5)

y = 0.8x + 1.2y