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Solve the linear programming problem by graphing. graph the feasible region, list the extreme points and identify the maximum value of Z. please list the equations of the lines that form the feasible region

Minimize z=4x+y

subject to

2x+4y>= 20

3x+2y<=24

x,y>=0

1 Answer

2 votes

Answer:

The minimum value of objective function is 5 at x=0 and y=5.

Explanation:

The given linear programming problem is

Minimize
z=4x+y

Subject to constraints


2x+4y\geq 20 .... (1)


3x+2y\leq 24 .... (2)


x,y\geq 0

The related line of both inequalities are solid lines because the sign of inequalities are ≤ and ≥. It means the points lie on related line are included in the solution set.

Check both inequalities by (0,0).


2(0)+4(0)\geq 20


0\geq 20

This statement is not true. So, the shaded region of inequality (1) will not contain the origin.


3(0)+2(0)\leq 24


0\leq 24

This statement is true. It means the shaded region of inequality (2) will contain the origin.


x,y\geq 0 means first quadrant.

The common shaded region is feasible region. The vertices of feasible region are (0,5), (0,12) and (7,1.5).

Calculate the value of objective function at these vertices.

For (0,5)


z=4(0)+(5)=5

For (0,12)


z=4(0)+(12)=12

For (7,1.5)


z=4(7)+(1.5)=29.5

Therefore the minimum value of objective function is 5 at x=0 and y=5.

Solve the linear programming problem by graphing. graph the feasible region, list-example-1
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