Please I need help with this question Describe a situation i

Please I need help with this question. Describe a situation in your own life or experience in which you would really like to maximize something, but you are limited by at least two constraints. Clearly describe both the objective to be maximized and the constraints. Could linear programming (section 7.5) be used to solve your problem? Why or why not? Minimum credit will be given to those whose problems cannot be solved using linear programming. Make sure that your problem can be solved using linear programming, show the actual equations with defined variables, your complete work, and the answer to your own question. ( please show work)

Solution

Real life situtaion : A dealer wants to purchase a combined total of no more than 200 refrigerators, and T.V. sets for inventory. Refrigerators weigh 200 pound each, and T.V. set weigh 50 pounds each. Suppose that the dealer has only a total of 10,000 pounds for these two items. If a profit of $35 for each refrigerator and $20 on eachT.V. set is projected, how many of each should be purchased and sold to make the largest profit?

Let x be the no of refrigerators

y be the n. of T.V. sets

x+ y < = 200

200x +50y < =10,000

The other two inequalities, x 0, y 0

are called non-negativity constraints and they are always added to the structural constraints in linear programming problems.

Objective Function : Profit functio is the objective function: P = 35x + 20y

Lets solve by simplex method.Form the tableau:

Tableau #1
x      y      s1     s2     s3     s4     p           
1      1      1      0      0      0      0      200  
200    50     0      1      0      0      0      10000
1      0      0      0      -1     0      0      0    
0      1      0      0      0      -1     0      0    
-35    -20    0      0      0      0      1      0    

Tableau #2
x      y      s1     s2     s3     s4     p           
1      1      1      0      0      0      0      200  
200    50     0      1      0      0      0      10000
-1     0      0      0      1      0      0      0    
0      1      0      0      0      -1     0      0    
-35    -20    0      0      0      0      1      0    

Tableau #3
x      y      s1     s2     s3     s4     p           
1      1      1      0      0      0      0      200  
200    50     0      1      0      0      0      10000
-1     0      0      0      1      0      0      0    
0      -1     0      0      0      1      0      0    
-35    -20    0      0      0      0      1      0    

Tableau #4
x      y      s1     s2     s3     s4     p           
0      0.75   1      -0.005 0      0      0      150  
1      0.25   0      0.005 0      0      0      50   
0      0.25   0      0.005 1      0      0      50   
0      -1     0      0      0      1      0      0    
0      -11.25 0      0.175 0      0      1      1750

Tableau #5
x      y      s1     s2     s3     s4     p           
0      0      1      -0.02 -3     0      0      0    
1      0      0      0      -1     0      0      0    
0      1      0      0.02   4      0      0      200  
0      0      0      0.02   4      1      0      200  
0      0      0      0.4    45     0      1      4000

Optimal Solution: p = 4000; x = 0, y = 200