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求出交點A(1,3)B(3,1)C(7,9) Z=X+2Y-4 ∴y=(-x/2)+(4+z)/2 ∴b=(4+z)/2 ∵b為直線y=(-x/2)+(4+z)/2的縱截距 ∴b越小,z越小 ∵過B b最小 ∴3+2-4-z=0 ∴Zmin=1
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