Delvecchios pizza shop makes three specialty pizzas the Migh

Delvecchio\'s pizza shop makes three specialty pizzas, the Mighty Meaty, the Very Veggie, and Super Cheesy. The Mighty Meaty is topped with 5 different meat toppings and 2 different cheeses. The Very Veggie has 6 different vegetable toppings and 4 different cheeses. The Super Cheesy has 6 different cheeses. The cost of t he meat toppings for each Mighty Meaty is $3, and the cost of the vegetable toppings is $2 fo r each Very Veggie. No more than $60 per day can be spent on these toppings. The cheese use d for the Mighty Meaty is $2 per pizza, the cheese for the Very Veggie is $4 per pizza, and the cheese for the Super Cheesy is $6 per pizza. No more than $480 per day can be spent on cheese. N o more than 80 pizzas can be made per day due to manpower and bake time. How many pizzas of each type should Delvecchio\'s pizza shop make in order to maximize revenue if the Might y Meaty sells for $15, the Very Veggie sells for $12, and the Super Cheesy sells for $12?

Solution

No. of Mighty Meaty, the Very Veggie, and Super Cheesy be x , y and z respectively

Toppings: Mighty Meat : 3*5*x

Ver veggie : 2*6*y

15x +12y <= $ 60 ----> 5x +4y <= 20 ---(1)

Cheese : Mighty meat: 2*x*2 = 4x

very veggie :4*4y= 16y

Super chees:6*6*z = 36z

So, 4x +16y +36z <= 480 ----> x+4y +9z <= 120 -----(2)

Total pizzas per day : x+y+z<=80 ----(3)

Further x>=0 ; y>=0 ; z>=0

Objective function: maximise revenue : 15x+12y+12z

Solving by simplex method:

Tableau #1
x y z s1 s2 s3 s4 s5 s6 p   
5 4 0 1 0 0 0 0 0 0 20   
1 4 9 0 1 0 0 0 0 0 120
1 1 1 0 0 1 0 0 0 0 80   
1 0 0 0 0 0 -1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-15 -12 -12 0 0 0 0 0 0 1 0

Tableau #2
x y z s1 s2 s3 s4 s5 s6 p   
5 4 0 1 0 0 0 0 0 0 20   
1 4 9 0 1 0 0 0 0 0 120
1 1 1 0 0 1 0 0 0 0 80   
-1 0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 -1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-15 -12 -12 0 0 0 0 0 0 1 0

Tableau #3
x y z s1 s2 s3 s4 s5 s6 p   
5 4 0 1 0 0 0 0 0 0 20   
1 4 9 0 1 0 0 0 0 0 120
1 1 1 0 0 1 0 0 0 0 80   
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 1 0 0 0 0 0 -1 0 0
-15 -12 -12 0 0 0 0 0 0 1 0

Tableau #4
x y z s1 s2 s3 s4 s5 s6 p   
5 4 0 1 0 0 0 0 0 0 20   
1 4 9 0 1 0 0 0 0 0 120
1 1 1 0 0 1 0 0 0 0 80   
-1 0 0 0 0 0 1 0 0 0 0
0 -1 0 0 0 0 0 1 0 0 0
0 0 -1 0 0 0 0 0 1 0 0
-15 -12 -12 0 0 0 0 0 0 1 0

Tableau #5
x y z s1 s2 s3 s4 s5 s6 p   
1 0.8 0 0.2 0 0 0 0 0 0 4
0 3.2 9 -0.2 1 0 0 0 0 0 116
0 0.2 1 -0.2 0 1 0 0 0 0 76   
0 0.8 0 0.2 0 0 1 0 0 0 4
0 -1 0 0 0 0 0 1 0 0 0
0 0 -1 0 0 0 0 0 1 0 0
0 0 -12 3 0 0 0 0 0 1 60   

Tableau #6
x y z s1 s2 s3 s4 s5 s6 p   
1 0.8 0 0.2 0 0 0 0 0 0 4
0 0.355556 1 -0.0222222 0.111111 0 0 0 0 0 12.8889
0 -0.155556 0 -0.177778 -0.111111 1 0 0 0 0 63.1111
0 0.8 0 0.2 0 0 1 0 0 0 4
0 -1 0 0 0 0 0 1 0 0 0
0 0.355556 0 -0.0222222 0.111111 0 0 0 1 0 12.8889
0 4.26667 0 2.73333 1.33333 0 0 0 0 1 214.667

Optimum soluition for the p= 15x+12y+12z occurs at x= 4 y =0 z = 12.889

maximum p = $ 214.667