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Help needed for simplex tableau problem (Solves by finding the E and E^-1 matrix)

Hi Guys,

I have this simplex problem which seems pretty difficult as it involves using the inverse matrix to figure out the optimum solution and we aren't allowed to follow the usual simplex tableau procedure. I am pretty stuck on it so I would appreciate any guidance you could give me.

The question says:

Jack is making a batch of smoothies to give to his friends. There are two fruits available: Banana and Orange. To make a nice smoothie Jack needs to make sure that at most 20% of the fruit in the smoothie is banana. One gram of banana costs 2p while one gram of orange costs 5p. He cannot buy all the oranges he likes because his local supermarket doesn't sell more than 400 grams to any customer.

The smoothie will also contain some vegetables: Carrots and Spinach. Any combination of vegetables is acceptable. Carrots cost 3p per gram and Spinach costs 4p. As he wants the smoothie to taste quite sweet, he does not want to use more than 800 grams of vegetables.

Jack also needs to include milk in the smoothie. He has 500 grams of milk which is about to go off so he decides to use all of it for his smoothies (this bit doesn't require a constraint).

To be able to feed all his friends, Jack needs to make at least 1500 grams of smoothie.

Jack has got part of the final tableau completed but he was too lazy to finish. Make an initial and complete the final tableau for him while finding the matrices E and E^-1 to find an optimal recipe for the smoothies and to determine how much it will cost him. You are not allowed to use the simplex method to answer the question.

End of Question.

I let x1 = banana, x2 = oranges, x3 = carrots and x4 = spinach

Objective Function: we want to minimise cost = -2x1 - 5x2 - 3x3 - 4x4

Constraints:

8x1 - 2x2 < 0 therefore 8x1 - 2x2 + s1 = 0 (For the 'no more than 20% of fruit is banana')
x2 < 400 therefore x2 + s2 = 0 (for the 'no more than 400g of oranges bought at shop')
x3 + x4 < 800 therefore x3 + x4 + s3 = 0 (for the 'no more than 800g of vegetables')
x1 + x2 + x3 + x4 > 1000 therefore x1 + x2 + x3 + x4 - s4 + A4 = 0 (for the final constraint about minimum weight. I reckon we subtract the milk weight from the 1500 grams total as it doesn't need its own constraint.)

I have attached a picture of what I think is the initial tableau but I think it might be wrong because I cant find the matrix E which has a determinate using the initial tableau that i created.

https://twitter.com/leborges23/status/933478830664187910

I have also attached a picture of what I have been able to do of the final tableau so far which would allow me to find the inverse of matrix E.

https://twitter.com/leborges23/status/933478891162931200

I know this is a big ask so don't worry if you don't have time. Any help would be appreciated because I just can't seem to find the correct E or E^1 by completing the final tableau.

Many thanks,
Alex