tolong bantu jawab no 12 pake cara

Buat persamaan 2 garisnya dahulu.
11x+3y = 30 (1)
5x+5,5y = 27,5 (2)

•A(0,5)
•B(x,y) =
eliminasi persamaan 1 dan 2
11x+3y = 30 |×5
5x+5,5y = 27,5 |×11

55x+15y = 150
55x+60,5y = 302.5
________________-
-45,5y = -152,5
y = 152,5÷45,5
y = 305÷91

subtitusi persamaan 1
11x+3y = 30
11x+3(305÷91) = 30
11x+(915÷91) = 30
11x = 30-(915÷91)
11 x = (1815÷91)
x= (1815÷91)÷11
x = (165÷91)

[x,y]
[(165÷91),(305÷91)]

•C[(5,5),0]

nilai persamaan 3x+4y

•A(0,5)
3x+4y
3(0)+4(5)
20

•B[(165÷91),(305÷91)]
3x+4y
3(165÷91)+4(305÷91)
(495÷91)+(1220÷91)
(245÷13)
18,84

•C[(5,5),0]
3x+4y
3(5,5)+4(0)
16,5+0
16,5

jadi nilai maksimum ada di titik A yaitu 20
jawabannya a. 20

semoga membantu :)
jangan lupa jawaban terbrainly sama terima kasihnya ya :)