3x1 + 6x2 greater than/equal to 18 Constraint #1

5x1 + 4x2 greater than/equal to 20 Constraint #2

8x1 + 2x2 greater than/equal to 16 Constraint #3

7x1 + 6x2 less than/equal to 42 Constraint #4

x1, x2 greater than/equal to 0 Non-negativity Constraint

(A) Optimal Solution (Graphical)

Constraint #1 3x1 + 6x2 =18

(0, 3) (6, 0)

Constraint #2 5x1 + 4x2 = 20

(0, 5) (4, 0)

Constraint #3 8x1 + 2x2 = 16

(0, 8) (2, 0)

Constraint #4 7x1 + 6x2 = 42

(0, 7) (6, 0)

Now by Plotting the Corner Points:

choose a point from the feasible region to plug into the objective function:

Let x1= 10 5(10) + 2x2 x2= -25

The optimal solution must be x1=2 and x2=3 because each constraint has been satisfied by these values. When we plug these values into the objective function we get 19 as our optimal result. The shaded region is the feasible region.

choose a point from the feasible region to plug into the objective function:

Let x1= 10 5(10) + 2x2 x2= -25

The optimal solution must be x1=2 and x2=3 because each constraint has been satisfied by these values. When we plug these values into the objective function we get 19 as our optimal result. The shaded region is the feasible region.

(B) Binding vs Non-Binding

Constraints numbers 2 and 3 are both binding, where as constraints 1 and 4 are not binding as they do not meet the eqality to the objective function and optimal solution as constaints 2 and 3 do. A constraint is binding as it approaches or touches the optimal solution/objective function.

(C) Slack vs Surplus Variables

(D) The feasible region has 3 corner/extreame points.

(E) New Objective Function

15x1 + 12x2-→ Minimize

choose 10 again: 15(10) + 12x2 x2= 12.5 x1= 8

Sales Price= 100 000/box

Variable Cost= 50 000/box

Fixed Cost= 1.5 million (1500000) dollars

(A) Profit function

z= (100 000x1) – (50 000x2) – 1 500 000

Break Even Analysis...