Question

PLEASE SOLVE!!

A wine manufacturer can source its grapes from two vineyards: vineyard A and vineyard B. Grapes from vineyard A costs $5 per ton to process, and grapes from vineyard B costs $10 per ton to process. Total processing cost must be kept to less than $80 per day. A previous contract with the vineyards require that the amount of grapes sourced from vineyard B cannot exceed twice the amount of grapes from vineyard A. In order to keep his plant running, the wine manufacturer must process at least 6 tons of grapes each day. Grapes from vineyard A yield 7.5 bottles of wine per ton, and grapes from vineyard B yield 20 bottles of wine per ton. How many tons of grapes from vineyard A and vineyard B must be processed each day to maximize the amount of wine produced subject to the above constraints? solve B, C, and D ONLY

(a) Formulate this problem as a linear program in the following 4 steps.

(a1) Identify the decision variables. Explain carefully what the decision variable (unknowns) represent.

x1 = tons of grapes from vineyard A

x2 = tons of grapes from vineyard B

(a2) Write out the objective function.

Max z = 7.5x1 + 20x2

For every ton of grapes from vineyard A you can make 7.5 bottles of wine. For every ton of grapes from vineyard B you can make 20 bottles of wine

(a3) Write out the constraints.

5x1 + 10x2 <= 80

x1 + x2 >= 6

-2x1 + x2 <= 0

x1 >= 0

x2 >= 0

(a4) Write the LP.

Max z = 7.5x1 + 20x2

5x1 + 10x2 <= 80

x1 + x2 >= 6

-2x1 + x2 <= 0

x1 >= 0

x2 >= 0

*****(b) Solve the LP. Draw the constraints, feasible region and 3 iso-profit lines on the blank graph below. Then indicate where the optimal solution is.  

*****(c) List all the corner points of the feasible region and calculate the value of the objective function at each of these points. Then say what the optimal objective value z* is.

Corner point	Objective function

*****(d) Suppose that grapes from vineyard A yield 10 bottles of wine per ton instead of 7.5, what is the optimal solution? (Hint: note that this will only change the objective function. Therefore, the feasible region and the corner points remain the same. List all the corner points of the feasible region and calculate the value of the new objective function at each of these points. Then say what the optimal objective value z* is.)

Corner point	Objective function

Answer 1

Points opoptimality are ( 6,0) (16,0) ( 3.2, 6.4) and (2.4)

Value of objective function  at these poonts

at (6,0) = 7.5x6 +0 = 45

at (16,0) = 7.5x16 = 120

at (3.2,6.4) = 3,2x7.5+20x6.4 =152

at ( 2,4) = 7.5x2+20x4 = 95

Max value of 152 is obtained at ( 3.2, 6.4)

(d)

If the new OF is 10x₁+20x₂

Value of objective function  at these poonts

at (6,0) = 10x6 +0 = 60

at (16,0) = 10x16 = 160

at (3.2,6.4) = 3.2x10+20x6.4 =160

at ( 2,4) = 10x2+20x4 = 100

Max value of 160 is obtained at two points ( 3.2, 6.4) and (16.0) both of which give optimal solution.