Min z = 80x_1 + 50x_2 s.t. 3x_1 + 1x_1 greaterthanorequalto 6 1x_1 +1x_2 greater

Question

Min z = 80x_1 + 50x_2 s.t. 3x_1 + 1x_1 greaterthanorequalto 6 1x_1 +1x_2 greaterthanorequalto 4 2x_1 + 6x_2 greaterthanorequalto 12 x_1, x_2 greaterthanorequalto 0 Optimal Obj. Value z = 353.33 Use Computer Output 1. Suppose the coefficient of x_2 in the objective function increases by 10. Which one of the following statements is guaranteed to be true? x_1 = 4, x_2 = 0.67, and z = 360.2 the dual price of constraint 3 will increase by 10 x_1 = 4, x_2 = 0.67, and z = 326.7 the z-value will be equal to 393.50

Explanation / Answer

z=80x1 + 60x2

which is the new objective function

The value of Z is bound to decrease, and dual prive od contraint 3 will not incraese by 10

Hence from the options it is clear that option c is the right answer.

Calculating it graphically also proves the solution

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