A cruise line is planning its menu for the next trip. Vacationers like eating steak, lobster, and chicken. The cruise line has decided to plan for at least forty percent of all booked passengers to have a steak dinner, for at least twenty percent of all passengers to have lobster, and the rest to have chicken. Steak dinners cost the company $10, lobsters cost $13, and chicken costs the line $7. On the next cruise, there are 400 passengers booked. In addition, the cruise line has decided to plan for an additional 25% more meals than bookings. Formulate the appropriate linear program.Let x1 = steak meals, x2 = lobster meals, x3 = chicken mealsMin. Z = 10x1 + 13x2 + 7x3S.t.x1 ≥ 160x2 ≥ 80-x1 -x2 +x3 ≥ 0x1 + x2 + x3 ≥ 500for all xi ≥ 0Upload your spreadsheet with the answer on sheet1How many steak lobster and chicken should they order?What was the total cost for the meals?

The cruiser will have 400 passangers on board. By taking into account that the cruise line decided to buy 25% more of meals than bookings, we can conclude that in order to calculate the ammount of meals they will order, we have to take the number of passangers on board, 400, and add to it 25% more. So the number of meal orders will be 400 + (1/4 * 400) = 500.

Let's use the letter S to simbolize the ammount of steak they should buy, with L the ammount of lobster, and with C the ammount of chicken. The sum of those ammounts should be 500, so we deduce this formula

[tex]S + L + C = 500[/tex]

we can replace C with 500 - S - L. Since at least forty percent of the passangers will have steak dinner and at least twenty percent will have lobster, we obtain this relations:

[tex]S \geq 400*0.4 = 160[/tex]
[tex]L \geq 400*0.2 = 80[/tex]

We note [tex]f[/tex] the total cost of the meals (in $). The value of [tex]f[/tex] will depend on the values of S and L. Using the information we are given about the cost of each dinner, and recalling that C was 500- S - L, we can represent [tex]f[/tex] by this formula

[tex]f(S,L) = 10*S + 13*L + (500-S-L) * 7[/tex]

Which is equal to [tex]10*S + 13*L + 3500 - 7*S - 7*L = 3500 + 3*S + 6*L[/tex]

We want to minimaze the value of [tex]f[/tex], since 3*S + 6*L will get bigger as we use bigger values of S and L, then we will be in a good position if we take the values of S and L as low as possible.

With the restriction we obtained, the lowest values of S and L we can take are 160 and 80 respectively; so we take those values. In conclusion

S = 160 (number of steak)
L = 80 (number of lobster)
C = 500-160-80 = 260 (number of chicken)

The total cost of the meals is

[tex]160*10 + 80*13 + 260*7 = 4460[/tex]