Question

Maximize P = 2x + 3y under the following constraints: 
 

x + 2y < 4

x + y < 3

x > 0, y > 0

Comments

i don't get it?

Answer 1

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

Now graph the all of four constraints.

•  Draw the coordinate plane.

Now first inequality x + 2y ≤ 4.

• Graph the line y = (- x/2) + 2
•  Since the inequality symbol is ≤ the boundary is included the solution set.
• Graph the boundary of the inequality y ≤ (- x/2) + 2 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

x + 2y ≤ 4

0 ≤ 4

The statement is true.

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

Similarly graph the other inequalities.

• Second inequality x + y ≤ 3

Test point (0, 0)

0  ≤ 3

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).

• Forth inequality y ≥ 0

Test point (1, 1)

1 ≥ 0

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

Graph

The feasible area looks like in the graph

To find maximum value we need to use corner point theorem.

From the graph the corner points are (0, 0) ,(0, 2),(2, 1),(3, 0)

The function P = 2x + 3y

Point	Function P = 2x + 3y	Value
(0, 0)	P = 2(0) + 3(0) = 0	0
(0, 2)	P = 2(0) + 3(2) = 6	6
(2, 1)	P = 2(2) + 3(1) = 7	7(Maximum)
(3, 0)	P = 2(3) + 3(0) = 6	6

The maximum value is 7 at (2,1).