Question

Solve the following LPP by using Simplex Method Max Z = 4X1 + 5X2 stc X1 + 2X2 ≤ 10 6X1 + 6X2 ≤ 36 X1 ≤ 4 And X1, X2 ≥ 0

Answer 1

Answer here is you answer

Explanation:

Answer 2

The solution is$X1 = 4, X2 = 2$ and $Z = 44 + 52 = 24$

To solve this linear programming problem (LPP) using the simplex method, we will first convert the inequalities into equations by introducing slack variables.

The problem becomes:

Max $Z = 4X1 + 5X2$

stc

X1 + 2X2 + S1 = 10

$6X1 + 6X2 + S2 = 36$

X1 + S3 = 4

And X1, X2, S1, S2, S3 ≥ 0

We will now construct the initial simplex tableau using the coefficients of the variables and the right-hand side values of the equations as follows:

X1 X2 S1 S2 S3 Z RHS

X1 1 2 1 6 1 4 10

X2 0 0 0 0 0 0 0

S1 0 0 0 0 0 0 10

S2 0 0 0 0 0 0 36

S3 1 0 0 0 0 0 4

Now we will apply the simplex method to solve the problem:

Pivot on X1 (pivot element 1 in row 1, column 1)

Divide Row 1 by pivot element:

| | X1 | X2 | S1 | S2 | S3 | Z | RHS |

|----|----|----|----|----|----|---|-----|

| X1 | 1 | 2 | 1 | 6 | 1 | 4 | 10 |

| X2 | 0 | -2 | -1 | -6 | -1 | -4| -20 |

| S1 | 0 | -1 | -1 | -3 | -1 | -2| -10 |

| S2 | 0 | 0 | 0 | 0 | 0 | 0 | 36 |

| S3 | 1 | 0 | 0 | 0 | 0 | 0 | 4 |

Subtract multiples of Row 1 from the other rows:

| | X1 | X2 | S1 | S2 | S3 | Z | RHS |

|----|----|----|----|----|----|---|-----|

| X1 | 1 | 2 | 1 | 6 | 1 | 4 | 10 |

| X2 | 0 | 0 | 0 | 0 | 0 | 0 | 16 |

| S1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

| S2 | 0 | 0 | 0 | 0 | 0 | 0 | 36 |

| S3 | 0 | 0 | 0 | 0 | 0 | 0 | 4 |

The solution is X1 = 4, X2 = 2.

$Z = 44 + 52 = 24$

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