CHAPTER 2 (DIFFRENTIATION)
GROUP 5
-MOHAMMAD AIMAN SYAZWAN BIN ARMAN (11DKM19F2003)
-MOHAMMAD NASEM MUSTAKIM BIN SAHRONI (11DKM19F2004)
A
2.5
5
7.5
10
12.5
15
V
45.8
-45.7
-79.1
-69.6
-14.2
70.1
V
100
80
y = 3.5263x2 - 59.133x + 167.7
60
40
20
0
0
-20
-40
-60
-80
-100
2
4
6
8
10
12
14
16
SOLUTION :
V = rA² + sAm + t
Y =3.5263x² - 59.133x +167.7
f’(x) = (2) 3.5263x - 59.133 + 0
= 7.0526x - 59.133 = 0
a=7.0526x
b=-59.133
c=0
7.0526x=59.133
x=59.133 / 7.0526
x1=8.385
When x1 = 8.385 y = 3.5263 (8.385)² - 59.133 (8.385) + 167.7
= -80.202
Stationary point ( 0 , 167.7 ) and ( 8.385 , - 80.202 )
F “ ( x ) = 7.0526 >0 , minimum point
Optimum point is (8.385,-80.202)