wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Solve the following Linear Programming Problems graphically:
Minimise Z=x+2y
subject to 2x+y3,x+2y6,x,y0.

Open in App
Solution


Given objective function is Z=x+2y

We have to minimize Z on constraints
2x+y3
x+2y6
x0,y0

After plotting the inequalities we got the feasible region as shown in the image

Now there are two corner points (0,3) and (6,0) lying on same line x+2y=6

Value at corner points are :

Corner Points Value of Z=x+2y
(0,3) 6 (minimum)
(6,0) 6 (minimum)
Since, feasible region is unbounded. So, 6 may or may not be minimum value.

Now to check if 6 is minimum or not, we have to draw Z<6x+2y<6

Since this region doesn't have any common region with feasible region.
So, 6 is the minimum value of Z.

810587_846970_ans_6942a7682983429595f1120bcf7e98b4.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Graphical Method of Solving LPP
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon