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Question 1176614: A school board is investigating various ways of composing faculty for a proposed
new elementary school. They can hire both teachers and aides. The amount of
money spent on salaries each year depends on how many teachers and aides are
hired. The board finds that the average teacher’s salary is $30,000 and the average
aide’s salary is $20,000. Suppose the board finds the following requirements
concerning the permissible number of teachers and aides:
The building can accommodate no more than 50 faculty members
A minimum of 20 faculty members is needed to staff the school
The school cannot be run entirely by aids: there must be at least 12 teachers
For a proper teacher-to-aide ratio, the number of teachers must be at least
two-thirds the number of aides.
It is impossible to hire a negative number of teachers or aides
Find the number of teachers and aides that could be hired that would minimize the
cost to the
district.
Objective Function (Cost): C =
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Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's set up and solve this linear programming problem.
**1. Define Variables**
* Let 't' be the number of teachers.
* Let 'a' be the number of aides.
**2. Formulate the Objective Function (Cost)**
The objective is to minimize the cost. The cost function (C) is:
* C = 30000t + 20000a
**3. Formulate the Constraints**
* **Total Faculty Constraint:** t + a ≤ 50
* **Minimum Faculty Constraint:** t + a ≥ 20
* **Minimum Teachers Constraint:** t ≥ 12
* **Teacher-to-Aide Ratio Constraint:** t ≥ (2/3)a
* **Non-negativity Constraints:** t ≥ 0, a ≥ 0
**4. Simplify the Teacher-to-Aide Ratio Constraint**
* 3t ≥ 2a
* a ≤ (3/2)t
**5. Find the Corner Points**
We need to graph the constraints and find the intersection points.
* **Constraint 1 (t + a ≤ 50):** a ≤ 50 - t
* **Constraint 2 (t + a ≥ 20):** a ≥ 20 - t
* **Constraint 3 (t ≥ 12):** Vertical line at t = 12
* **Constraint 4 (a ≤ (3/2)t):** a ≤ 1.5t
Let's find the intersections:
* **Intersection of t = 12 and a = 20 - t:** a = 20 - 12 = 8. Point (12, 8)
* **Intersection of t = 12 and a = 1.5t:** a = 1.5 * 12 = 18. Point (12, 18)
* **Intersection of t = 12 and a = 50 - t:** a = 50 - 12 = 38. Point (12, 38)
* **Intersection of a = 1.5t and a = 50 - t:** 1.5t = 50 - t, 2.5t = 50, t = 20, a = 1.5 * 20 = 30. Point (20, 30)
* **Intersection of a = 20 - t and a = 1.5t:** 20 - t = 1.5t, 20 = 2.5t, t = 8. This violates t>=12. Therefore, this intersection is not in the feasible region.
* **Intersection of a = 20 - t and a = 50 - t:** Not possible, as they are parallel.
The corner points of the feasible region are:
* (12, 8)
* (12, 18)
* (12, 38)
* (20, 30)
**6. Evaluate the Objective Function at Each Corner Point**
* **(12, 8):** C = 30000(12) + 20000(8) = 360000 + 160000 = 520000
* **(12, 18):** C = 30000(12) + 20000(18) = 360000 + 360000 = 720000
* **(12, 38):** C = 30000(12) + 20000(38) = 360000 + 760000 = 1120000
* **(20, 30):** C = 30000(20) + 20000(30) = 600000 + 600000 = 1200000
**7. Determine the Minimum Cost**
The minimum cost is $520,000, which occurs at the point (12, 8).
**Answers**
* **Teachers:** 12
* **Aides:** 8
* **Objective Function (Cost):** C = 30000t + 20000a
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