00:01
The problem says that we need to maximize p is equals to x plus 2 times y and it is subjected to 30x plus 20y less than equals to 600 .1x plus 0 .4y less than equals to 4 and 0 .2x plus 0 .3y less than equals to 4 and 0 .2x plus 0 .3y less.
00:34
Less than equals to 4 .5 and we have the non -negative restriction which is x greater than equals to 0 and y greater than equals to 0 now first of all let's draw these equations on a graph paper to find the feasible reason so substitute y is equal to 0 we get x is equal to 20 and when we we substitute x is equals to 0, we get y is equals to 30.
01:17
Similarly, when we substitute y is equal to 0, we get x is equal to 40, and when we substitute x is equals to 0, we get y is equals to 10.
01:34
Now similarly when we substitute y is equals to 0, we get x is equals to 22 .5 comma, and when we substitute x is equals to 0 we get y is equals to 15 now based on these coordinates let's draw the graph for these equations and the graph is shown below so here we have the graph for given three equations this red line represents 30x plus 20 y is equals to 600 this green line represents 0 .2x plus 0 .3y is equal to 4 .5 and this blue line represents 0 .1x plus 0 .4y less than equals to 4.
02:38
Here we also have less than equals to and also less than equals to.
02:43
Now from this graph we can observe that this one is the visible common region.
02:55
This one and in this feasible reason we have five points first one is this this is second this is third this is fourth and this one is fifth and from the graph it is clear that the coordinates of this point is 0 comma 0 coordinates of this point is 20 comma 0 coordinates of this point is 0 .10 now let's find the coordinates of this point of this point and this point...