36) Consider the following minimization problem:

Min z = x1 + 2x2
s.t. x1 + x2 ≥ 300
2x1 + x2 ≥ 400
2x1 + 5x2 ≤ 750
x1, x2 ≥ 0

What is the optimal solution?

In the real or integer domain?

If integer domain, have you done linear programming yet?
If so, graphics solution or simplex?

To find the optimal solution, you can use the simplex method to solve the linear programming problem.

Step 1: Convert the problem into its standard form. Since all the constraints are inequalities, we need to introduce slack variables to convert them into equations.

Let s1, s2, and s3 be the slack variables for the three constraints, respectively.

The problem becomes:

Minimize z = x1 + 2x2

subject to:

x1 + x2 + s1 = 300
2x1 + x2 + s2 = 400
2x1 + 5x2 - s3 = 750

x1, x2, s1, s2, s3 ≥ 0

Step 2: Construct the initial simplex tableau.

1 | 2 | 0 | 0 | 0 | 300
2 | 1 | 1 | 0 | 0 | 400
2 | 5 | 0 | -1 | 0 | 750
-1 | -2 | 0 | 0 | 1 | 0

Step 3: Select the most negative value in the bottom row, which is -2. This corresponds to the variable x2.

Step 4: Divide each element in the second row by the pivot (-2), and update the tableau.

-0.5 | -1 | 0 | 0.5 | 0 | -200
2 | 1 | 1 | 0 | 0 | 400
-1.5 | 0 | 0 | 0.5 | 0 | 950
-0.5 | 0 | 0 | 1 | 0 | 200

Step 5: Select the most negative value in the bottom row, which is -1.5. This corresponds to the variable x1.

Step 6: Divide each element in the third row by the pivot (-1.5), and update the tableau.

1 | 2/3 | 0 | -1/3 | 0 | 900/3
2 | 1 | 1 | 0 | 0 | 400
1 | 0 | 0 | -1/3 | 0 | 200/3
1 | 0 | 0 | -2/3 | 0 | 600/3

Step 7: Select the most negative value in the bottom row, which is -2/3. This corresponds to the variable s3.

Step 8: Divide each element in the fourth row by the pivot (-2/3), and update the tableau.

1 | 2/3 | 0 | -1/3 | 0 | 900/3
2 | 1 | 1 | 0 | 0 | 400
1 | 0 | 0 | -1/3 | 0 | 200/3
-1.5 | 0 | 0 | 1 | 0 | -600/3

Step 9: All numbers in the bottom row are non-negative, so the current tableau is the final tableau.

The optimal solution is:
x1 = 200/3
x2 = 400/3
s1 = 0
s2 = 0
s3 = 0

Therefore, the optimal solution is x1 = 200/3 and x2 = 400/3.

To find the optimal solution for this minimization problem, we can use linear programming techniques. Here's how we can solve it step by step:

Step 1: Formulate the problem
The given problem can be written in the standard form of a linear programming problem as follows:

Minimize:
z = x1 + 2x2

Subject to:
x1 + x2 ≥ 300 -- (Constraint 1)
2x1 + x2 ≥ 400 -- (Constraint 2)
2x1 + 5x2 ≤ 750 -- (Constraint 3)
x1, x2 ≥ 0

Step 2: Graph the feasible region
To determine the feasible region, we graph the three inequality constraints. The feasible region is the region that satisfies all the constraints.

Constraint 1: x1 + x2 ≥ 300
This constraint represents the line x1 + x2 = 300. To graph it, plot the points (0, 300) and (300, 0), and draw a line connecting these two points.

Constraint 2: 2x1 + x2 ≥ 400
This constraint represents the line 2x1 + x2 = 400. To graph it, plot the points (0, 400) and (200, 200), and draw a line connecting these two points.

Constraint 3: 2x1 + 5x2 ≤ 750
This constraint represents the line 2x1 + 5x2 = 750. To graph it, plot the points (0, 150) and (375, 0), and draw a line connecting these two points.

Shade the region that is bounded by all three lines. This shaded region represents the feasible region.

Step 3: Identify the corner points of the feasible region
The feasible region is the intersection of the shaded region and the non-negative axes (x1, x2 ≥ 0). To find the corner points of the feasible region, locate the points where the lines forming the feasible region intersect.

Step 4: Calculate the objective function at each corner point
Evaluate the objective function (z = x1 + 2x2) at each corner point of the feasible region. The corner point that produces the smallest value of z will be the optimal solution.

Step 5: Determine the optimal solution
Compare the values of z at each corner point. The corner point with the smallest value of z is the optimal solution.

Note: If there are no intersection points or if the feasible region is unbounded, the linear programming problem may not have an optimal solution.

By following these steps, you can determine the optimal solution for the given minimization problem using linear programming techniques.