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can you please tell me the steps involved in making the basic simplex tables for this question. ?

Maximize: Z = 15x1 +25x2

subject to : 5x1 +5x2 < equal to 25

10x1 + 15x2 < equal to 60

x1,x2 > equal to 0

(note - x is a variable. not the multiplication symbol)

thanks.

Sorry for the delay.

We take the coefficients, negate the Z row, and look for the greatest negative in the Z row.

We also add a slack variable for each of the equation.

The start of it is:

-15 -25

5 5 1 0 25

10 15 0 1 60

The entering variable would be x2, since -25 is more negative than -15.

This makes the column for x2 be the pivot column.

Now for the 1st equation, 25/5 = 5, and for the 2nd equation, 60/15 = 4.

Since 4 is smaller, so the 2nd equation is the pivot row.

This makes the 15 be the pivot element.

Since that's the pivot element, add 5/3 times the 2nd row to the Z row,

add -1/3 times the 2nd equation to the 1st equation,

and divide the 2nd equation by 15.

After only one iteration, that gives us

1.67 0 0 1.67 100

1.67 0 1 -0.33 5

0.67 1 0 0.07 4

Since there are no more negatives in the Z row, the optimum value has been achieved.

Differential Equations

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