Question

Soal cerita turunan, mohon bantuannya ya

Answer 1

1. B(x) = 2x²-800x+200rb

B(x) = 2.(x²-400x+100rb)

maks ketika B'(x) = 0

B'(x) = 2.(2x-400) = 0

2x-400 = 0

x-200 = 0

x = 200 barang

Keuntungan maks : B(200)

B(200) = 2.(200²-400.200+100rb)

           = 400(200-400+500)

           = 400.(200+100)

           = 400.300

B(200) = Rp.120.000,-

2. L = p.l

L = (5x+5)(5-x)

L = -5(x+1)(x-5)

L = -5(x²-4x-5)

maksimum ketika L' = 0

L' = -5.(2x-4) = 0

2x-4 = 0

x-2 = 0

x = 2

p = 5(x+1) = 5.(2+1) = 15

l = -(x-5) = -(2-5) = -(-3) = 3

L max = 15.3 = 45 cm²

3. x+y = 40

x = 40-y

misalkan xy di subtitusikan m

m = xy

m = (40-y).y

m = 40y-y²

maksimum ketika m' = 0

m' = 0 = 40 - 2y

2y = 40

y = 20

x = 40-y = 40-20 = 20