Nilai maksimum z = 3x + 4y , apabila x dan y merupakan titik pada daerah penyelesaian sistem pertidaksamaan 2x + 3y ≤ 18; 6x + 5y ≤ 38; x ≥ 0; y ≥ 0 adalah …

[tex]z = 3x + 4y \\ 2x + 3y \leqslant 18 \\ x = 9 \\ y = 6 \\ 6x + 5y \leqslant 38 \\ x = \frac{38}{6} = 6.3 \\ y = \frac{38}{5} = 7.5 \\ \\ a = 3(6.3) + 4(0) = 18.9 \\ b = 3(0) + 4(6) = 24 \\ c = 3(3) + 4(4) = 25 \\ \\ 2x + 3y = 18( \times 3) \\ 6x + 9y = 54 - 6x + 5y = 38 \\ 4y = 16 \\ y = 4 \\ 2x + 3y = 18 \\ 2x + 3(4) = 18 \\ 2x = 6 \\ x = 3[/tex]