   Chapter 4.3, Problem 11E

Chapter
Section
Textbook Problem

In Exercises 5 − 20 , use the method of this section to solve each linear programming problem. Maximize                     P = x + 2 y subject   to                   2 x + 3 y ≤ 12                                           − x + 3 y = 3                                             x ≥ 0 , y ≥ 0

To determine

To solve:

The given linear programming problem.

Maximize     P=x+2y

subjectto    2x+3y12          x+3y=3           x0,y0

Explanation

Approach:

1. In the standard maximization problem the objective function is to be maximized.

2. All variable involved in the problem should be non-negative.

3. Each linear constraint may be written so that the expression involving the variables is less than or equal to a non-negative constraint.

4. The optimal solution is reached when all the entries of bottom row of the simplex tableau are non-negative.

Given:

The given problem is,

Maximize     P=x+2y

subjectto    2x+3y12          x+3y=3           x0,y0

Calculation:

Consider the given linear programming problem.

The constraint x+3y=3 has two constraints,

x+3y3x+3y3

Change the sign for all as .

x+3y3x3y3

So, now the problem is,

Maximize     P=x+2y

subjectto    2x+3y12          x+3y3           x3y3           x0,y0

Introduce the slack variables u, v, and w. in the given linear programming problem,

2x+3y+u    =12x+3y   +v   =3x3y    +w=3x2y      +P=0

The initial simplex tableau is,

 x y u v w P Constant 2 3 1 0 0 0 12 −1 (3) 0 1 0 0 3 1 −3 0 0 1 0 −3 −1 −2 0 0 0 1 0

Table (1)

At x=0,y=0 we have u=12,v=3,w=3,P=0

In the initial simplex tableau the minimum value for bottom row is 2 so pivot this column, The pivoted element is 3 so now 13R2

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