Ma Thematic

Anindya Sela Pratiwi Gilda Miranda Uli Rianiari Xi science 2 SMAN 8 Pekanbaru

• English • Indonesian

1. If x + y = 82, find the maximum value of xy !

ANSWER

1.Jika x + y = 82, carilah nilai maksimal dari xy !

JAWABAN

L(x) = xy = x(82-x) = 82x-x² L’(x) = 82-2x 0 = 82-2x 82 = 2x x = 41 y = 41 So,

xy = 1681 NEXT

L(x) = xy = x(82-x) = 82x-x² L’(x) = 82-2x 0 = 82-2x 82 = 2x x = 41 y = 41 xy = 1681 SELANJUTNYA

2. Find the minimum value of the sum of a positive number and its reciprocal !

ANSWER

2.Carilah nilai minimal dari jumlah angka positif dan kebalikannya ! JAWABAN

1 =x + x x 2 +1 = x L ' =0 2 x ( x ) −( x 2 +1)(1) =0 2 x 2 x 2 −x 2 −1 =0 x 2 =1 x =±1

NEXT

1 = x+ x 1 = 1+ 1 =2 NEXT

1 =x + x 2 x +1 = x L' =0 2 x ( x ) −( x 2 +1)(1) =0 2 x 2 x 2 −x 2 −1 =0 x 2 =1 x = ±1

SELANJUTNYA

1 = x+ x 1 = 1+ 1 =2 SELANJUTNYA

3. A rectangular field is surrounded by a fence on three of its sides and a straight hedge on the fourth side. If the length of the fence is 320 m, find the maximum area of the field enclosed! ANSWER

3.Daerah persegi panjang dikelilingi oleh pagar pada 3 sisinya dan pagar tanaman lurus pada sisi keempat. Jika panjang pagarnya 320m, carilah luas maksimal dari daerah yang tertutup ! JAWABAN

2 x + y = 320 y = 320 − 2 x x

y

x

L = x. y L = x(320 − 2 x) L = 320 x − 2 x 2 L' = 0 320 − 4 x = 0 320 = 4 x x = 80 NEXT

y = 320 − 2 x y = 320 − 2(80) y = 160 Max Area

= x. y = 160.80 = 12800 HOME

2 x + y = 320 y = 320 − 2 x x

y

L = x. y L = x(320 − 2 x) L = 320 x − 2 x 2

x

L' = 0 320 − 4 x = 0 320 = 4 x x = 80 SELANJUTNYA

y = 320 − 2 x y = 320 − 2(80) y = 160 Luas Max

= x. y = 160.80 = 12800 KEMBALI