Question

Dari soal matematika ini tolong berikan jawaban nomor4&5

Answer 1

4. model 1 =x   model 2=y
x+2y<=20................(i)

3/2x+1/2y<=10 ......(ii)
3x+y<=20

  x+2y<=20
6x+2y<=40
----------------   ---
-5x     <= -20
         x<=4

x+2y<=20
4+2y<=20
    2y<=16
      y<=8

jadi total pakian maksimum jadi model 1 + model 2 = 4+8=12

5. tas a =x    tas b =y

x+y=25 ...... (1)

30.000x+40.000y <= 840.000
3x+4y<=84 .............. (2)

3x+3y<=75
3x+4y<=84
-------------  ------
   y     <= 9

x+y <= 25
x+9 <=25
 x    <=16

jadi keuntungan maksimum x + y= 16(5000) + 9(8000)= 80.000 +72.000 =152.000