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Words 431

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DECISION VARIABLES

Let x1= number of jars of chow-chow relish to produce

Let x2 = number of jars of tomato relish to produce

OBJECTIVE FUNCTION

Zmax=$2.25x1 + $1.95x2

Subject to

8x1 + 6x2 ≤ 1920 ounces…………………Cabbage constraint

3x1 + 6x2 ≤ 1440 ounces…………………Tomato constraint

3x1 + 2x2 ≤ 720 ounces…………………..Onion constraint

X1 + x2 ≤ 288 jars………………………...Total Production constraint

X1,x2 ≥ 0………………………………… non-negativity

| | CONSTRAINTS | ounces | | | CABBAGE | TOMATO | ONIONS | PROFIT $ | RESOURCES | | | | | CHOW-CHOW | 8 | 3 | 3 | 2.25 | TOMATO | 6 | 6 | 2 | 1.95 | AVAILABLE | 1920 | 1440 | 720 | |

Q1(b)

Using the graph,

8x1 + 6x2 = 1920 ounces…………………………. (1)

3x1 + 6x2 = 1440 ounces…………………………. (2)

3x1 + 2x2 = 720 ounces…………………………… (3)

X1 + x2 =288 jars…………………………………….. (4)

If x1 in equation (1) =0, then

8(0) + 6x2 =1920

6x2=1920

X2=1920/6

X2=320

(0, 320)

If x2 in equation (1) =0, then

8x1 + 6(0) = 1920

X1=1920/8

X1=240

(240, 0)

If x1 in equation (2) =0, then

3(0) + 6x2 = 1440

6x2 = 1440

X2 = 1440/6

X2 = 240 (0, 240)

If x2 in equation (2) = 0, then

3x1 + 6(0) = 1440

3x1 = 1440

X1 = 1440/3

X1= 480 (480, 0)

If x1 in equation (3) =0, then

3(0) + 2x2 = 720

2x2 =720

X2=720/2

X2=360 (0, 360)

If x2 in equation (3) =0, then

3x1 + 2(0) =720

3x1 = 720

X1 =720 3

X1 = 240 (240, 0)

If x1 in equation (4) = 0, then

X2=288

(0, 288)

If x2 in equation (4) = 0, then

X1 = 288 (288, 0)

Due to the multiple constraints, it is difficult to obtain the optimal solution from the graph. Therefore, the simultaneous equation would be used to the solve linear programming model.

Using simultaneous equation,

8x1 + 6x2 = 1920 ounces…………………………. (1)

3x1 + 6x2 = 1440 ounces…………………………. (2)

3x1 + 2x2 = 720 ounces…………………………… (3)

X1 + x2 =288 jars………………………………….. (4)

Using the...

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