Question

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

Answer 1

Answer:

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

Step-by-step 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.