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Consider the linear programming problem :

Minimize z=50x+70yz = 50x + 70y

Subject to 2x+y≥82x+y \geq 8, x+2y≥10x+2y \geq 10, x≥0x \geq 0, y≥0y \geq 0

The minimum value of objective function is :

Solution

✅ Correct Option: 1

Corner points: (0,8)(0,8) gives z=560z=560; (10,0)(10,0) gives z=500z=500; intersection of 2x+y=82x+y=8 and x+2y=10x+2y=10 is (2,4)(2,4) giving z=100+280=380z=100+280=380. Minimum is 380380.

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