00:01
So it is given that the maximize of this question is going to maximize 23x plus 19y.
00:09
This objective function is to maximize the value of 23x plus 19y with respect to subject to constraints are given three constraints.
00:19
X plus 5y is equal to 20.
00:25
8x minus 2y is equal to 5 and 5x plus y is less than or equal to 75.
00:35
From first equation what we can get if x equals to zero then y equals 4 and if y equals zero x equals 20.
00:46
So this is the first optimal solution from equation one.
00:52
For second if x equals to zero then y equals to minus of 5 by 2 and if y equals to zero x equals to 5 by 8.
01:04
The second option and for third equation we'll get it y equals to zero x equals to 75 by 5 and that is equal to 15 and if x equals to zero we have y equals to 75.
01:24
From this we can draw the figure and from which we can find the optimal solution.
01:29
The highest value of this is nearly equals 86 or 48, 28 to the 8, 0 .625.
01:40
So highest value of y is we can see to 75.
01:46
75 here y is 0, here y is 0, minus 5 by 2 or minus 2 .5.
01:53
So y value is also minus 2 .5.
01:56
This we can draw.
01:58
So first we can see if x equals to zero y equals to 4.
02:02
Also y equals to 4.
02:05
So x equals to zero y equals to 4.
02:06
This is first one and y equals to zero x equals to 20.
02:11
X equals to 20, x equals to 0 .625, 0 .625, x equals 20 and also x equals 15.
02:23
So we can just now draw y equals 4 and x equals 0, x equals 20.
02:29
So this is the first line which we can draw this one.
02:35
Second is x equals zero y equals to minus 2 .5.
02:38
So this one x equals to minus 2 .5 and x equals to 0 .625 and y equals to 0 .625.
02:44
This is the second line which we can draw.
02:49
This is 0 .625 and last one y equals to x equals 15.
02:53
So this is 1 .75...