Question

solve the following L.L.P where maximum Z = 2000 x + 3000y ,
subject to
3x +3y <= 36 ; 5x +2y <= 80 ; 2x + 6y <=60
x >= 0 ; y >= 0​

Answer 1

Answer:

for y = 9 & x = 3

Max Z = 33000

Step-by-step explanation:

maximum Z = 2000 x + 3000y

Max value will be with max value of Y & max value of x + y together

3x +3y <= 36

x + y <= 12

x <= 12

y <= 12

2x + 6y <= 60

x + 3y <= 30

for max x = 12 y <=6  & min x = 0 y <= 10

5x + 2y <= 80

for x = 12 y<= 10  and for y = 12  x <= 11.2

From all these equation we can conclude that max value of y satisfying all conditions

if y = 10

x + y <= 12 will give x <=2

5x + 2y <=80 will give x<=12

x + 3y <= 30 will give x =0

min value of x = 0 satisfying all conditions

x + y = 10

if y = 9

x + y <= 12 will give x <=3

5x + 2y <=80 will give x<=12

x + 3y <= 30 will give x <=3

min value of x = 3 satisfying all conditions

x + y = 12

Z = 2000 * 3 + 3000*9

=> Z = 6000 + 27000

=> Z = 33000