DB45
Tabel kain 1 = x kain II = y ........x...y...persediaan A.....2...4.......20 ----> 2x+ 4y ≤20 atau x + 2y ≤10 B ....2...2.......14 -----> 2x+2y ≤14 atau x + y ≤ 7 f(x,y) = 40.000 x + 50.000 y . titik potong x + 2y = 10 x + y = 7 --------------(-) y = 3 --> x= 7- 3= 4 tikpotong(x,y) = (4,3)
grafiknya daerah penyelesaian di lampiran x+ 2y = 10 jika x= 0, y = 5 --> (0,5) jika y=0, x = 10 -->(10,0
x + y = 7 jika x= 0 , y = 7 --> (0,7) jika y= 0 , x = 7 --> (7,0) ..... daerah tertutup dengan titik pojok nya di (0,5) --> f(xy) = 5(50.000) = 250.000 (4.3) --> f(x,y)= 4(40.000)+ 3(50.000) = 310.000 (7,0) --> f(x,y) = 7(40.000) = 280.000
kain 1 = x
kain II = y
........x...y...persediaan
A.....2...4.......20 ----> 2x+ 4y ≤20 atau x + 2y ≤10
B ....2...2.......14 -----> 2x+2y ≤14 atau x + y ≤ 7
f(x,y) = 40.000 x + 50.000 y
.
titik potong
x + 2y = 10
x + y = 7
--------------(-)
y = 3 --> x= 7- 3= 4
tikpotong(x,y) = (4,3)
grafiknya daerah penyelesaian di lampiran
x+ 2y = 10
jika x= 0, y = 5 --> (0,5)
jika y=0, x = 10 -->(10,0
x + y = 7
jika x= 0 , y = 7 --> (0,7)
jika y= 0 , x = 7 --> (7,0)
.....
daerah tertutup dengan titik pojok nya di
(0,5) --> f(xy) = 5(50.000) = 250.000
(4.3) --> f(x,y)= 4(40.000)+ 3(50.000) = 310.000
(7,0) --> f(x,y) = 7(40.000) = 280.000
pendapatan maksimum = 310.000