The Erlanger Manufacturing Company makes Coffee Makers and Irons. The profit est

Question

The Erlanger Manufacturing Company makes Coffee Makers and Irons. The profit estimates are $5 for each coffee maker sold and $10 for each iron. The labor-hour requirements for the products in the three production departments are department A-coffee makers 0.50 and irons-3.00, B-coffee makers 2.00 and irons 1.00, and C-coffee makers 0.25 and irons 0.25. The departments production supervisors estimate that the following number of labor-hours will be available during the next month: 450 hours in department A, 350 hours in department B, and 50 hours in department C.

a. Develop a linear programming model to maximize profits.

b. Find the optimal solution. How much of each product should be produced, and what is the project profit?

c. What are the scheduled production time and slack time (i.e., available time after production) in each department?

Explanation / Answer

Let coffee makers = M and Irons = I

max 5M + 10I

constraints are

.5M + 3I <= 450 (department A)

2M + 1 I <= 350 (department B)

.25M + .25I <= 50 (department C)

M, I >= 0

Note that all our constraints are <=, coefficients are positive, and the coefficients of the optimization function are positive, so we know that our solution will be on the boundary.

Letting I = 0, we have .5M <= 450; 2M <= 350, .25M <= 50

Solving, M <= 450/.5 = 900 M <= 350/2=175 M <= 50/.25 = 200

Thus, we have a boundary point (175, 0)

This is from line 2M + 1 I <= 350

Letting M = 0, we have 3I <= 450; I <= 350; .25I <= 50

Solving, I <= 450/3 = 150; I <= 150; I <= 50/.25 = 200

Thus, we have boundary point (0, 150)

This is from line .5M + 3I <= 450

Find where

.5M + 3I = 450

2M + 1 I = 350

Multiplying the bottom equation by 3

6M + 3I = 1050

Subtracting the top equation from this,

5.5M = 600

M = 1200/11

2M + I = 350

2400/11 + I = 350

I = 350 - 2400/11

I = 1450/11

The boundary point is (1200/11, 1450/11)

Note that, at this intersection, .25M + .25I = 1/4(1200/11) + 1/4(1450/11) = 1/4(2650/11) = 1325/22

This is greater than 50.

Thus, where these lines intersect is not part of the boundary.

Thus, we must consider all three lines as part of the boundary.

.5M+3I = 450

.25M+.25I = 50

Double the bottom equation

.5M + .5I = 100

Subtract.

2.5I = 350

I = 350/2.5 = 140

.25M = 50 - .25I = 50 - .25(140) = 15

M = 15/.25 = 60

Thus, we have (60, 140) on the boundary.

.25M+.25I = 50

2M + 1 I = 350

Multiplying the top equation by 4,

M + I = 200

Subtract from the bottom equation.

M = 150

I = 350 - 2M = 350 - 2(150) = 350 - 300 = 50

The intersection is (150, 50)

Thus, we have four boundary points, (175, 0), (150, 50), (60, 140), (0, 150)

Consider the optimality functions at these four points.

(175,0) 5M + 10I = 5(175) + 10(0) = 875

(150, 50) 5M + 10I = 5(150) + 10(50) = 750 + 250 = 1000

(60, 140) 5M + 10I = 5(60) + 10(140) = 300 + 1400 = 1700

(0, 150) 5M + 10I = 5(0) + 10(150) = 1500

The optimal solution is 1700 at (60, 140), or M = 60 and I = 140

This was at the intersection of

.5M+3I = 450

.25M+.25I = 50

Thus, there is no slack for department A and department C

For Department B,

2M + I = 2(60) + 140 = 120 + 140 = 260

Thus, we schedule Departments A, B, and C as follows: A: 450 hours; B: 260 hours; C: 50 hours

The constraintfor B is 2M + 1 I <= 350

350 - 260 = 90

We have 90 hours slack time for Department B and no slack time for departments A and C.

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