Question

Solve the linear programming problem. Minimize and maximize Upper P equals negative 20 x plus 30 y Subject to 2 x plus 3 y greater than or equals 30 2 x plus y less than or equals 26 negative 2 x plus 3 y less than or equals 30 x comma y greater than or equals 0

Answer 1

Answer:

Maximum = 540 at (6,14)

Minimum = 300 at (0,10) or (12,2).

Step-by-step explanation:

The given linear programming problem is

Minimize and maximize: P = 20x + 30y

Subject to constraint,

$2x+3y\ge 30$            .... (1)

$2x+y\le 26$            .... (2)

$-2x+3y\le 30$            .... (3)

$x,y\geq 0$

The related equation of given inequalities are

$2x+3y=30$

$2x+y=26$

$-2x+3y=30$

Table of values are:

For inequality (1).

x      y

0     10

15     0

For inequality (2).

x      y

0     26

13     0

For inequality (3).

x      y

0     10

15     0

Pot these ordered pairs on a coordinate plane and connect them draw the corresponding related line.

Check each inequality by (0,0).

$2(0)+3(0)\ge 30\Rightarrow 0\ge 30$    False

$2(0)+(0)\le 26\Rightarrow 0\le 26$     True

$-2(0)+3(0)\le 30\Rightarrow 0\le 30$    True

It means (0,0) is included in the shaded region of inequality (2) and (3), and (0,0) is not included in the shaded region of inequality (1).

From the below graph it is clear that the vertices of feasible region are (0,10), (6,14) and (12,2).

Calculate the values of objective function on vertices of feasible region.

Point           P = 20x + 30y

(0,10)           P = 20(0) + 30(10) = 300

(6,14)           P = 20(6) + 30(14) = 540

(12,2)           P = 20(12) + 30(2) = 300

It means objective function is maximum at (6,14) and minimum at (0,10) or (12,2).