00:01
Hi, now we are going to find the maximum value of z.
00:05
Now the given z is 4x1 plus 6x2 and subject to the constraints minus x1 plus x2 less than or equal to 11, x1 plus x2 less than or equal to 27, 2x1 plus 5x2 less than or equal to 90.
00:26
Now i write these inequalities as minus x1 plus x2 is equal to 11 and i label this one as equation number 1, x1 plus x2 is equal to 27 and i label this one as equation number 2, 2x1 plus 5x2 is equal to 90 and i label this one as equation number 3.
00:49
Now in the first equation if i put x1 is equal to 0 then we get the value of x2 is 11.
00:58
If i put x2 is equal to 0 then we get the value of x1 is minus 11.
01:04
So the points of equation 1 are 0, 11 and minus 11, 0 and in the second equation if i put x1 is equal to 0 then we get the value of x2 is 27 and if i put x2 is equal to 0 then we get the value of x1 is equal to 27.
01:32
So the points of equation 2 are 0, 27 and 27, 0 and next in the third equation if i put x1 is equal to 0 then we get the value of x2 is 18.
01:51
If i put x2 is equal to 0 then we get the value of x1 is 45.
01:58
So the points of equation 3 are 0, 18 and 45, 0 and next we are going to draw these three lines in the graph then we get this graph.
02:17
Now here this is the feasible region.
02:21
Now we have to find the intersection of the first and third equation and the intersection of the second and third equation.
02:34
Now from the first equation we can write y is equal to 11 plus x.
02:39
Now i am going to substitute this value of y in the third equation then we get 7x is equal to 35.
02:48
From this we get the value of x is 5 then the value of y will be 11 plus 5 that is 16...