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Solution to Z = 4x1 6x2 x1 x2 ≤ 4; 3x1 x2 ≤ 12; x1, x2 ≥ 0 is: Ops: A. Unique B. Unbounded C. Degenerate D. Infinite?
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Solution to Z = 4x1 6x2 x1 x2 ≤ 4; 3x1 x2 ≤ 12; x1, x2 ≥ 0 is: O...
Solution to linear programming problem

Objective function:
Z = 4x1 + 6x2

Constraints:
x1 + x2 ≤ 4
3x1 + x2 ≤ 12
x1, x2 ≥ 0

Step 1: Plotting the constraints

To solve the linear programming problem graphically, we need to plot the constraint lines on the coordinate plane.

x1 + x2 ≤ 4 can be rewritten as x2 ≤ -x1 + 4, which gives us the line:

x2 = -x1 + 4

Similarly, 3x1 + x2 ≤ 12 can be rewritten as x2 ≤ -3x1 + 12, which gives us the line:

x2 = -3x1 + 12

Plotting these two lines on the coordinate plane, we get the following graph:



The feasible region is the shaded area that satisfies all the constraints.

Step 2: Finding the corner points

To find the optimal solution, we need to evaluate the objective function at each corner point of the feasible region. The corner points are the vertices of the feasible region.

In this case, the corner points are:

(0, 0), (0, 4), (2, 2), and (4, 0)

Step 3: Evaluating the objective function

We evaluate the objective function Z = 4x1 + 6x2 at each corner point:

Z(0, 0) = 0
Z(0, 4) = 24
Z(2, 2) = 20
Z(4, 0) = 16

Step 4: Conclusion

The optimal solution is Z = 24 at (0, 4).

Answer: The solution to the linear programming problem is unique.
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Solution to Z = 4x1 6x2 x1 x2 ≤ 4; 3x1 x2 ≤ 12; x1, x2 ≥ 0 is: Ops: A. Unique B. Unbounded C. Degenerate D. Infinite?
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