MINIMIZATION BY THE SIMPLEX METHOD

Answer:

1 Unit of Food A

7 Unit of Food B

1 Unit of Food C

Explanation:

There are 3 Types of Foods: A, B, C

Costs of these Foods are $3, $2, $3 respectively.

Let the person consume \(x, y, z\) units of the Foods respectively.

Therefore, Total Cost i.e \(W=3x+2y+3z\)

In this Question, we have to minimize \(Z\).

Now, we will make conditions for minimization as per given data in the question.

\(x+2y+2z\ge 8\) ....Taking Vitamins of each Food

\(x+y+z\ge 9\) ....Taking Minerals of each Food

\(2x+y+2z\ge 10\) ....Taking Calories of each Food

Additionally, it is given that at least 1 unit of each food needs to be purchased. So,

\(x,y,z\ge 1\)

Now, Forming the Tableau and solving the problem:

Tableau #1
x y z s1 s2 s3 s4 s5 s6 -Z
1 2 2 -1 0 0 0 0 0 0 8
1 1 1 0 -1 0 0 0 0 0 9
2 1 2 0 0 -1 0 0 0 0 10
1 0 0 0 0 0 -1 0 0 0 1
0 1 0 0 0 0 0 -1 0 0 1
0 0 1 0 0 0 0 0 -1 0 1
3 2 3 0 0 0 0 0 0 1 0

Tableau #2
x y z s1 s2 s3 s4 s5 s6 -Z
1 0 2 -1 0 0 0 2 0 0 6
1 0 1 0 -1 0 0 1 0 0 8
2 0 2 0 0 -1 0 1 0 0 9
1 0 0 0 0 0 -1 0 0 0 1
0 1 0 0 0 0 0 -1 0 0 1
0 0 1 0 0 0 0 0 -1 0 1
3 0 3 0 0 0 0 2 0 1 -2

Tableau #3
x y z s1 s2 s3 s4 s5 s6 -Z
1 0 0 -1 0 0 0 2 2 0 4
1 0 0 0 -1 0 0 1 1 0 7
2 0 0 0 0 -1 0 1 2 0 7
1 0 0 0 0 0 -1 0 0 0 1
0 1 0 0 0 0 0 -1 0 0 1
0 0 1 0 0 0 0 0 -1 0 1
3 0 0 0 0 0 0 2 3 1 -5

Tableau #4
x y z s1 s2 s3 s4 s5 s6 -Z
0.5 0 0 -0.5 0 0 0 1 1 0 2
0.5 0 0 0.5 -1 0 0 0 0 0 5
1.5 0 0 0.5 0 -1 0 0 1 0 5
1 0 0 0 0 0 -1 0 0 0 1
0.5 1 0 -0.5 0 0 0 0 1 0 3
0 0 1 0 0 0 0 0 -1 0 1
2 0 0 1 0 0 0 0 1 1 -9

Tableau #5
x y z s1 s2 s3 s4 s5 s6 -Z
0 0 0 -0.5 0 0 0.5 1 1 0 1.5
0 0 0 0.5 -1 0 0.5 0 0 0 4.5
0 0 0 0.5 0 -1 1.5 0 1 0 3.5
1 0 0 0 0 0 -1 0 0 0 1
0 1 0 -0.5 0 0 0.5 0 1 0 2.5
0 0 1 0 0 0 0 0 -1 0 1
0 0 0 1 0 0 2 0 1 1 -11

Tableau #6
x y z s1 s2 s3 s4 s5 s6 -Z
0 0 0 0 0 -1 2 1 2 0 5
0 0 0 0 -1 1 -1 0 -1 0 1
0 0 0 1 0 -2 3 0 2 0 7
1 0 0 0 0 0 -1 0 0 0 1
0 1 0 0 0 -1 2 0 2 0 6
0 0 1 0 0 0 0 0 -1 0 1
0 0 0 0 0 2 -1 0 -1 1 -18

Tableau #7
x y z s1 s2 s3 s4 s5 s6 -Z
0 0 0 0 -1 0 1 1 1 0 6
0 0 0 0 -1 1 -1 0 -1 0 1
0 0 0 1 -2 0 1 0 0 0 9
1 0 0 0 0 0 -1 0 0 0 1
0 1 0 0 -1 0 1 0 1 0 7
0 0 1 0 0 0 0 0 -1 0 1
0 0 0 0 2 0 1 0 1 1 -20

Therefore,

Optimal Solution: w = 20; x = 1, y = 7, z = 1

Therefore:

1 Unit of Food A

7 Unit of Food B

1 Unit of Food C

Does that help?