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,
[tex]2x+3y\ge 30[/tex] .... (1)
[tex]2x+y\le 26[/tex] .... (2)
[tex]-2x+3y\le 30[/tex] .... (3)
[tex]x,y\geq 0[/tex]
The related equation of given inequalities are
[tex]2x+3y=30[/tex]
[tex]2x+y=26[/tex]
[tex]-2x+3y=30[/tex]
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).
[tex]2(0)+3(0)\ge 30\Rightarrow 0\ge 30[/tex] False
[tex]2(0)+(0)\le 26\Rightarrow 0\le 26[/tex] True
[tex]-2(0)+3(0)\le 30\Rightarrow 0\le 30[/tex] 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).
