$Maximize Z = 5 x_{1} + 3 x_{2},$

$Subject to :$

$x_{1} + 2 x_{2} \leq 10$

$x_{1} - x_{2} \leq 8$

$x_{2}, x_{2} \geq 0$

In the starting simplex tableau, $x_{1} and x_{2}$ are non-basic variables and the value of Z is zero. The value of Z in the next simplex tableau is .
40

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Solution

The correct option is A 40

Objective function will increase by 5 per unit of $x_{1}$ and bring 8 units.

$So, total increase (8 \times 5 = 40)$

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Similar questions

P = 10 x_{1} + 5 x_{2}

4 x_{1} + x_{2} \leq 28

2 x_{1} + 3 x_{2} \leq 24

x_{1}, x_{2} \geq 0

- x_{1} + 3 x_{2} \leq 10 x_{1} + x_{2} \leq 6 x_{1} - x_{2} \leq 2 x_{1}, x_{2} \geq 0

P = 70 x_{1} + 50 x_{2}

4 x_{1} + 3 x_{2} \leq 240

2 x_{1} + x_{2} \leq 100

x_{1}, x_{2} \geq 0

x_{1}, x_{2}

5 x^{2} + b x - 28 = 0 & 5 x_{1} + 2 x_{2} = 1 \forall b \in Z .

b .

Consider the primal problem in LP.

Maximize Z = 4 X_{1} + 3 X_{2}

Subject to

X_{1} + X_{2} \leq 8 2 X_{1} + X_{2} \leq 10 X_{1}; X_{2} \geq 0

together with its dual (LD) Then

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