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)