Consider the following linear programming problem maximize z...

Consider the following linear programming problem. Maximize Z = 4x1 + 2x2, subject to 6x1 ≤ 24 (resource 1) 1x1 + 7x2 ≤ 37 (resource 2) 1x2 ≤ 13 (resource 3) and x1 ≥ 0, x2 ≥ 0 Solve this problem graphically and find an optimal solution. The optimal solution is ( ,..... ,...... ) with Z = .......

Transcript

00:01 So in this problem, we are asked to solve for the maximum z is 4x1 plus 2x2.

00:12 So we have two resources, x1 and x2.

00:16 We're trying to maximize them here.

00:18 Subject to these constraints.

00:28 Okay.

00:29 I have 6x1 is less than or equal to 24.

00:36 I have x1 plus 7x2 is less than.

00:41 Than are equal to 37.

00:44 I have x2 is less than or equal to 13.

00:49 X1 is greater than equal to zero, and x2 is greater than equal to zero.

00:55 Okay.

00:56 So we're asked to solve this graphically and determine the values, of course, for x1 and x2, which will write like that.

01:06 Okay.

01:08 So we're going to make a little table here, x1, x2, and z.

01:15 Because we're going to make a little table here.

01:16 It's going to make a little table here.

01:16 We're trying to find the maximum z here.

01:19 And we need to graph all of the constraints.

01:24 Notice that this one, if i divide by 6, means that x1 is less than or equal to 4.

01:31 Okay.

01:32 So i went to desmos .com and brought down the graphing calculator here.

01:39 And so i do the first one where i will say that x is less than or equal to or gives me that zone right there doesn't it gives me that area there the next one we say that x plus y so in other words i'm making x2 be y oh that's supposed to be 7y 7y is less than or equal to 37 okay so now i'm in this zone aren't i okay what's next we have that x2 which is y for us is less than or equal to 13...

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