Let’s see:

Z = 5X1 + 3X2

With constraints:

X1 + X2 ≤ 6

X1 ≥ 3

X2 ≥ 3

2X1 + 3X2 ≥ 3

Additional constraints (that are true but logically unnecessary):

X1 ≥ 0 ; X2 ≥ 0

Personally, I find the first three constraints a little odd. Both X1 and X2 can only equal 3. There are no other values that can possibly satisfy:

X1 + X2 ≤ 6

Where did you get the problem?

Tom ]]>

then it become x1 + x2 = 6

x1=3

x2=3

2×1+3×2 = 3

After substitution in equations to find the points

(3,3) is the point pf interaction of x1=3 and x2=3

and the corner points in the feasible region is (3,0) , (5,0) .

then substitution in the objective function z = 5×1 + 3×2 .

the maximum value at x=5 and x2=0

maximun value is 25

the minimum value at x1=3 and x2=0 .

minimum value is 15

i’am i right ??

could any one guide me . ]]>

Consider the graphical representation of the following LPP

Maximize

Z = 5×1

+ 3 x2

Subject to x1 +

x2 6,

x1 3

x2

3

2×1 + 3 x2

3

x1 0 , x2 0.

Answer the following questions:

(a) In each of the following cases indicate if the

solution space ( the feasible region) has one point, infinite number of points, or no points.

1.

The constraints are

as given above.

2.

The constraint x1

+ x2 6

is changed to x1 + x2 5.

3.

The constraint x1

+ x2 6

is changed to x1 + x2 7.

(b) For each case in (a) , determine the number of feasible extreme points, if any.

(c) For

the cases in (a) in which a feasible solution

exists , determine the maximum values of Z .

(d) Answer part (c) for minimize Z.

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