SOLUTION: The director of MIS Center At a large company wants to do stuff consulting station with two shifting teams: Team A will compromise of 3 senior programmers and 3 system analyst and

Question 1185780: The director of MIS Center At a large company wants to do stuff consulting station with two shifting teams: Team A will compromise of 3 senior programmers and 3 system analyst and team B will consist 2 senior programmers and 5 system analysts. the director wants to use no more than 42 individuals.
There will be at least 48 hours to be filed during the week, with team A serving for 4 hours and the team B shift servings for 3 hours. The cost of team A is 3200 per hour and 2800 per hour for team B shift.
Determine the number of shifts each team has to render in order to minimize the cost.
•What is the Test ratio for table 2?
•What is the pivotal/al column in Tableu 1?
•What is the result in the last row of Tableu 4?
•Show the proper representation of Tableu

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
This is a linear programming problem. Here's how to set it up and solve it:
**1. Define Variables:**
* Let *x* be the number of shifts Team A works.
* Let *y* be the number of shifts Team B works.
**2. Objective Function (Minimize Cost):**
Cost = 3200 * 4 * x + 2800 * 3 * y = 12800x + 8400y
**3. Constraints:**
* **Personnel Constraint:** 6x + 7y ≤ 42 (Team A has 6 members, Team B has 7)
* **Hours Constraint:** 4x + 3y ≥ 48
* **Non-negativity Constraints:** x ≥ 0, y ≥ 0 (Can't have negative shifts)
**4. Graphical Solution:**
The most straightforward way to solve this with these numbers is graphically.
1. **Plot the constraints:** Treat each inequality as an equation and plot the lines on a graph. Shade the feasible region that satisfies all the inequalities.
2. **Identify the corner points:** The optimal solution will always lie at one of the corner points of the feasible region. The corner points are the intersections of the constraint lines. Find the coordinates of these points.
3. **Evaluate the objective function:** Plug the x and y coordinates of each corner point into the cost function (12800x + 8400y).
4. **Optimal Solution:** The corner point that yields the *lowest* cost is the optimal solution.
**Example of Finding Corner Points:**
* Intersection of 6x + 7y = 42 and 4x + 3y = 48:
Solve this system of equations (substitution or elimination). One way to do it is to multiply the first equation by 2 and the second equation by 3:
12x + 14y = 84
12x + 9y = 144
Subtracting the first from the second gives 5y = 60, so y = 12. Plugging back into either equation gives x = -7.5. Since x and y must be non-negative we can ignore this.
Intersection of 6x + 7y = 42 and x = 0: x = 0 gives y = 6.
Intersection of 4x + 3y = 48 and y = 0: y = 0 gives x = 12.
Intersection of 4x + 3y = 48 and 6x + 7y = 42:
Multiply the first equation by 7 and the second by 3:
28x + 21y = 336
18x + 21y = 126
Subtracting the second from the first gives 10x = 210, so x = 21. Plugging back into either equation gives y = -20. Since x and y must be non-negative we can ignore this.
The corner points are (0, 6) and (12, 0).
**Example of Evaluating the Objective Function:**
* At (0, 6): Cost = 12800(0) + 8400(6) = 50400
* At (12, 0): Cost = 12800(12) + 8400(0) = 153600
In this example, the minimum cost is 50400 when Team A works 0 shifts and Team B works 6 shifts.
**Simplex Method:**
The Simplex Method is generally used for problems with more than two variables. Since this problem only has two variables, the graphical method is much more efficient. However, here's how you'd set up the initial tableau if you *had* to use the Simplex Method:
1. **Convert inequalities to equations:** Introduce slack and surplus variables.
6x + 7y + s1 = 42
4x + 3y - s2 + a1 = 48 (a1 is an artificial variable)
x, y, s1, s2, a1 ≥ 0
2. **Initial Tableau:**
The initial tableau would be set up with the coefficients of the variables, including the objective function (which you would set equal to zero and move to the other side: -12800x - 8400y + C = 0). The artificial variable requires a special treatment in the objective function (often a large positive number "M").
Because the graphical method is so much easier for this two-variable problem, I highly recommend using that approach. It's much faster and less prone to calculation errors.

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.

The analysis in the post by @CPhill is either incomplete or incorrect.

Below I explain WHY it is so.

If to accept interpretation of the problem as in the post by @CPhill, then the problem is maximize 12800x + 8400y under these constraints 6x + 7y <= 42 4x + 3y >= 48 x >= 0, y >= 0. But the feasibility domain under these constraints is . Indeed, it is obvious that in positive domain for x and y 4x + 3y < 6x + 7y. So, if (x0,y0) is the point in the feasibility domain, then 48 <= 4x0 + 3y0 < 6x0 + 7y0 <= 42, and we arrive to the absurdist inequality 48 < 42. This contradiction disproves the existence of solutions. So, the answer for this analysis is that the given problem HAS NO solution.

        There is EMPTINESS at this point.

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