Consider the following linear program: Min s. t. 8X + 12Y 1X + 3Y >= 9 2X + 2Y >= 10 6X + 2Y >= 18 A, B >= 0 a. Use the graphical solution procedure to find the optimal solution. b. Assume that the objective function coefficient for X changes from 8 to 6. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution. C. Assume that the objective function coefficient for X remains 8, but the objective function coefficient for Y changes from 12 to 6. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution. d. The computer solution for the linear program in part (a) provides the following objective

Graph of (A) (B) and {D) are attached accordingly.

Explanation:

A)

The critical region of the constraints can be seen in the following diagram -

(0,9) (0,5) (0,3) (0,0) (3,0) (5,0) (9,0) The feasible region is shown in white

The intersection points are found by using these equations -

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 48

(9,0) x+3y = 9; y = 0 72

(2,3) 2x+2y = 10; 6x+2y = 18 52

(0,9) 6x+2y = 18; x = 0 108

So, we can see the minimum value of the objective function occurs at point (3,2) and the minimum value of the objective function is = 48.

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B)

When we change the coefficients of the variables in the objective function, the optimal solution may or may not change as the weights (coefficient) are different for each constraints for both the variabls. So, it all depends on the coefficient of the variables in the constraints.

In this case, the optimal solution does not change on changing the coefficient of X from 8 to 6 in the objective function.

The critical region would remain same (as shown below) as it is defined by the constraints and not the objective function.

(0,9) (0,5) (0,3) (0,0) (3,0) (5,0) (9,0) The feasible region is shown in white

However, the optimal value of the objective function would change as shown below-

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 42

(9,0) x+3y = 9; y = 0 54

(2,3) 2x+2y = 10; 6x+2y = 18 48

(0,9) 6x+2y = 18; x = 0 108

So, we can see that the minimum value now has become 42 (which had to change obviously).

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C)

Now, when we change the coefficient of the variable Y from 12 to 6, again the critical region would remain same as earlier. But in this case, the optimal solution changes as shown below -

Vertex Lines Through Vertex Value of Objective

(3,2) x+3y = 9; 2x+2y = 10 36

(9,0) x+3y = 9; y = 0 72

(2,3) 2x+2y = 10; 6x+2y = 18 34

(0,9) 6x+2y = 18; x = 0 54

We can see that the minimum value now occurs at (2,3) which is 34, so both the optimal solution and optimal value have changed in this case.

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D)

When we limit the range of the variables as -

4 \leq X \leq 8 \:\: and\:\: 12\leq Y \leq 24,

the critical region now becomes -

So, the new critical points are (4,12), (4,24), (8,24) and (8,12).

So, the values of the objective function at these points can be calculated as -

Vertex Value of Objective

(4,12) 8*4+12*12 = 176

(4,24) 8*4+12*24 = 320

(8,24) 8*8+12*24 = 352

(8,12) 8*8+12*12 = 208

So, the new optimal solution is (4,12) and the optimal value is 176.

if we knew the range of the variables in the part B and C earlier, we could have just said that the optimal solution will not change as the value would have been no longer depended on the coefficients of variables in the constraints.

answer; institution based;