Question

(1 point) Use the graphical method to maximize P = 6x1 + 4x2 subject to x1 2x1 x1 + + + x2 3x2 2x2 > 11 30 5 22 x120 x2 > 0 If there are no solutions, enter DNE in each box. Maximum value is P = where x1 = and x2 =

Answer 1

Problem is

MAX P(zx)=6x1+4x2

subject to:

x1+x2≥11

2x1+3x2≥30

x1+2x2≤22

and x1,x2≥0;

Hint to draw constraints

1. To draw constraint

x1+x2≥11→(1)

Treat it as x1+x2=11

When x1=0 then x2=?

⇒(0)+x2=11

⇒x2=11

When x2=0 then x1=?

⇒x1+(0)=11

⇒x1=11

x1	0	11
x2	11	0

2. To draw constraint

2x1+3x2≥30→(2)

Treat it as 2x1+3x2=30

When x1=0 then x2=?

⇒2(0)+3x2=30

⇒3x2=30

⇒x2=30/3=10

When x2=0 then x1=?

⇒2x1+3(0)=30

⇒2x1=30

⇒x1=30/2=15

x1	0	15
x2	10	0

3. To draw constraint

x1+2x2≤22→(3)

Treat it as x1+2x2=22

When x1=0 then x2=?

⇒(0)+2x2=22

⇒2x2=22

⇒x2=22/2=11

When x2=0 then x1=?

⇒x1+2(0)=22

⇒x1=22

x1	0	22
x2	11	0

The value of the objective function at each of these extreme points is as follows:

Extreme Point Lines through Extreme Point Objective function value

Coordinates z=6x1+4x2

Extreme Point	Lines through Extreme Point	Objective function value
Coordinates		z=6x1+4x2
(x1,x2)
A(0,11)	1→x1+x2≥11	6(0)+4(11)=44
	3→x1+2x2≤22
B(22,0)	3→x1+2x2≤22	6(22)+4(0)=132
	5→x2≥0
C(15,0)	2→2x1+3x2≥30	6(15)+4(0)=90
	5→x2≥0
D(3,8)	1→x1+x2≥11	6(3)+4(8)=50
	2→2x1+3x2≥30

The graphical solution of the same is as follows:

Graphical Solution II =< ZX+ IX X1+ 2x2<= 22 2x 14 3x2 >= 30 x y1 Oy

The maximum value of the objective function z=132 occurs at the extreme point (22,0).

Hence, the optimal solution to the given LP problem is : x1=22,x2=0 and max z=132.

Hence P = 132,

X1 = 22,

X2 = 0