Question

1. Kilgore’s Deli is a small delicatessen located near a major university. Kilgore’s does a large walk-in carry-out lunch business. The deli offers two luncheon chili specials, Wimpy and Dial 911. At the beginning of the day, Kilgore needs to decide how much of each special to make (he always sells out of whatever he makes). The profit on one serving of Wimpy is $0.4, on one serving of Dial 911, $0.53. Each serving of Wimpy requires 0.2 pound of beef, 0.2 cup of onions, and 5 ounces of Kilgore’s special sauce. Each serving of Dial 911 requires 0.2 pound of beef, 0.35 cup of onions, 2 ounces of Kilgore’s special sauce, and 5 ounces of hot sauce. Today, Kilgore has 15 pounds of beef, 10 cups of onions, 84 ounces of Kilgore’s special sauce, and 55 ounces of hot sauce on hand.

   1. Develop a linear programming model that will tell Kilgore how many servings of Wimpy and Dial 911 to make in order to maximize his profit today. If required, round your answers to two decimal places. For subtractive or negative numbers use a minus sign even if there is a + sign before the blank. (Example: -300)
      

      Let	W = number of servings of Wimpy to make
      	D = number of servings of Dial 911 to make

      Max	W	+	D
      s.t.
      	W	+	D			(Beef)
      	W	+	D			(Onions)
      	W	+	D			(Special Sauce)
      	W	+	D			(Hot Sauce)
      			W, D	≥	0

   2. Find an optimal solution. If required, round your answers to two decimal places.
      
      W =________ , D =________ , Profit = ______$  
      
   3. What is the shadow price for special sauce? If required, round your answers to two decimal places.
      
      $ _________
      
   4. Increase the amount of special sauce available by 1 ounce and re-solve. If required, round your answers to two decimal places.
      
      W =______ , D =_________ , Profit = ______$  
      
      Does the solution confirm the answer to part (c)?
      

Answer 1

(a)

Maximize 0.4W + 0.53D

s. t.

0.2W + 0.2D ≤ 15

0.2W + 0.35D ≤ 10

5W + 2D ≤ 84

0W + 5D ≤ 55

W, D ≥ 0

(b) Solver solution

Solution: W = 12.40, D = 11.00, Profit = $10.79

(c)

Constraints
			Final	Shadow	Constraint	Allowable	Allowable
	Cell	Name	Value	Price	R.H. Side	Increase	Decrease
	$B$5	Beef	4.68	0	15	1E+30	10.32
	$B$6	Onions	6.33	0	10	1E+30	3.67
	$B$7	Special sauce	84	0.08	84	91.75	62
	$B$8	Hot sauce	55	0.074	55	67.96296296	55

Shadow price for special sauce = $0.08

(d) Solver solution

Solution: W = 12.60, D = 11.00, Profit = $10.87

Yes, profit in (c) has increased by $0.08 in (d)

[Please give me a Thumbs Up if you are satisfied with my answer. If you are not, please comment on it, so I can edit the answer. Thanks.]