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The Apex Television Company has to decide on the number of 60- and 42- inch sets

ID: 3043445 • Letter: T

Question

The Apex Television Company has to decide on the number of 60- and 42- inch sets to be produces at one of its factories. Market research indicates that at most 40 of the 60-inch sets and 10 of the 42-inch sets ca be sold per month. The maximum number of workhours available is 500 per month. A 60-inch set required 20 work-hours and 42-inch set requires 10 work-hours. Each 60-inch set sold produces a profit of $120 and each 42-inch set produces a profit of $80. A wholesaler has agreed to purchase all the television sets produces if the numbers do not exceed the maxima indicated by the market research. Formulate a linear programming model for this problem and find the maximum profit of the company. Define the Variable. Formulate an OPL.

60-inch TV’s = x

42-inch TV’s = y

x 40, maximum 60-inch TV’s that can be produced.

y 10, maximum 42-inch TV’s that can be produced.

Based on the work hours per month:

20x + 10y 500 (available work hours)

10y 500 – 20x

y 50-2x

The profit function is:

Z= 120x + 80y = 3200

(0,0) = 0

(25,0) = 3,000

(0,10) = 800

(20,10) = 3,200

(20,10) is the maximum due to a $3,200 profit. They should produce 20 of the 60- inch TV’s and 10 of the 42-inch TV’s.

Explanation / Answer

60-inch TV’s = x

42-inch TV’s = y

x 40, maximum 60-inch TV’s that can be produced.

y 10, maximum 42-inch TV’s that can be produced.

Based on the work hours per month:

20x + 10y 500 (available work hours)

The profit function is:

Z= 120x + 80y

Tableau #1
x      y      s1     s2     s3     p           
1      0      1      0      0      0      40   
0      1      0      1      0      0      10   
20     10     0      0      1      0      500  
-120   -80    0      0      0      1      0    

Tableau #2
x      y      s1     s2     s3     p           
0      -0.5   1      0      -0.05 0      15   
0      1      0      1      0      0      10   
1      0.5    0      0      0.05   0      25   
0      -20    0      0      6      1      3000

Tableau #3
x      y      s1     s2     s3     p           
0      0      1      0.5    -0.05 0      20   
0      1      0      1      0      0      10   
1      0      0      -0.5   0.05   0      20   
0      0      0      20     6      1      3200

Optimal Solution: p = 3200; x = 20, y = 10