Maximize z=15x+10y subject to:

Question

3x+2y less than or equal to 54

4x+5y less than or equal to 100

x greater than equal to 0, y greater than to 0

The linear programming problem has:

a.) unique solution

b.) multiple solution

c.) unbounded solution

d.) no feasible solution

Answer 1

The inequalities are 3x + 2y ≤ 54, 4x + 5y ≤ 100, x ≥ 0 and y ≥ 0

Now graph the all of four constraints.

•  Draw the coordinate plane.

Now first inequality 3x + 2y ≤ 54.

• Graph the line y = (- 3x/2) + 27
•  Since the inequality symbol is ≤ the boundary is included the solution set.
• Graph the boundary of the inequality 3x + 2y ≤ 54 with solid line.
• To determine which half plane to be shaded use a test point in either half- plane.
• A simple choice is (0, 0). Substitute x = 0 and y = 0 in original inequality

3x + 2y ≤ 54

0 ≤ 54

The statement is true.

• Since the statement is true , shade the region contain point (0,0).

Similarly graph the other inequalities.

• Second inequality 4x + 5y ≤ 100

Test point (0, 0)

0  ≤ 100

Since the statement is true , shade the region contain point (0, 0).

• Third inequality x ≥ 0

Test point (1, 1)

1 ≥ 0

Since the statement is true , shade the region contain point (1, 1).

• Fourth inequality y ≥ 0

Test point (1, 1)

1 ≥ 0

Since the statement is true , shade the region contain point (1, 1).

Graph

 

Answer 2

Contd...

The feasible area looks like in the graph

From the graph the corner points are (0, 0) ,(0, 20),(10, 12),(18, 0)

The function z = 15x + 10y

Point	Function z = 15x + 10y	Value
(0, 0)	z = 15(0) + 10(0) = 0	0
(0, 20)	z = 15(0) + 10(20) = 200	200
(10, 12)	z = 15(10) + 10(20) = 350	350(Maximum)
(18, 0)	z = 15(18) + 10(0) = 270	270

The maximum value is 350 at (10,12).

Optimal solution at x = 10, y = 20.

Option a is correct.