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Jawab
x + 4y = 120 ⇒ titik A (0,30) B(120,0)
x + 4 = 60 ⇒ titik C (0,60) D(60, 0)
x + 4y = 120
x + 4 = 60
3y = 60
y = 60/3
y = 20
x + 4y = 120
x + 4(20) = 120
x + 80 = 120
x = 120 - 80
x = 40
titik E (40, 20)
titik pojok daerah HP = {(0,0)(0,30)(40,20)(60,0)}
Nilai (f, x) = 10x + 20y
Nilai (0,0) ⇒ 0
Nilai (0,30) ⇒ 20(30) = 600
Nilai (40,20) ⇒ 10(40) + 20(20) = 800
Nilai (60,0) ⇒ 10(60) + 0(0) = 600
Jadi, nilai maksimumnya adalah 600