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x =0 --> 2y ≤ 12
y ≤ 6 ----> (0,6)
y =0 --> x ≤ 12 --> (12, 0)
2x + y ≤ 12
x = 0, y ≤12 ---> (0,12)
y = 0, 2x ≤ 12
x ≤ 6 ----> (6,0)
titik ptong kedua garis :
x+ 2y = 12 I x 2 ---> 2x + 4y = 24
2x + y = 12 2x + y = 12
---------------------- -
3y = 12
y = 4
x + 2(4) = 12
x = 4 - ---> (4,4)
nilai maks f(x,y)
f(0,6)= 2.0² + 3(0)(6) + 6²
= 36
f(6,0) = 2.6² + 3(6)(0) + 0²
= 72
f(4,4) = 2.4² + 3(4)(4) + 4²
= 32 + 48 + 16
= 96
sehingga nilai maksimum ada di f(4,4) = 96