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Question-212651




Question Number 212651 by hardmath last updated on 20/Oct/24
Answered by Ghisom last updated on 20/Oct/24
x+3y+5z≤15  x+y+z≤7  2x+y+4z≤12  ========  x+3y+5z≤15  y+2z≤4  5y+6z≤18  ========  x+3y+5z≤15  y+2z≤4  z≤(1/2)  ========  x≤(7/2)  y≤3  z≤(1/2)  ========  2x+3y+z≤((33)/2)
$${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15} \\ $$$${x}+{y}+{z}\leqslant\mathrm{7} \\ $$$$\mathrm{2}{x}+{y}+\mathrm{4}{z}\leqslant\mathrm{12} \\ $$$$======== \\ $$$${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15} \\ $$$${y}+\mathrm{2}{z}\leqslant\mathrm{4} \\ $$$$\mathrm{5}{y}+\mathrm{6}{z}\leqslant\mathrm{18} \\ $$$$======== \\ $$$${x}+\mathrm{3}{y}+\mathrm{5}{z}\leqslant\mathrm{15} \\ $$$${y}+\mathrm{2}{z}\leqslant\mathrm{4} \\ $$$${z}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$======== \\ $$$${x}\leqslant\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${y}\leqslant\mathrm{3} \\ $$$${z}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$======== \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}+{z}\leqslant\frac{\mathrm{33}}{\mathrm{2}} \\ $$

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