Max z = 4x1 + 3x2 subject to constraints

2x1 + x2 ≤ 1000

x1 + x2 ≤ 800

x1 ≤ 400

x2 ≤ 700

x1, x2 ≥ 0

To maximize the profit ‘z’, find the optimum units of x1 & x2 to be used.

1
x1 = 100, x2 = 700
2
x1 = 400, x2 = 200
3
x1 = 200, x2 = 600
4
x1 = 300, x­2 = 500

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