A small firm builds two types of products type a requires 20...

A small firm builds two types of products. Type A requires 20 minutes on threading machine, 10 minutes on slotting machine and 7.5 minutes of craftsman time. Type B requires 10 minutes on threading machine, 20 minutes on slotting machine and 30 minutes of craftsman time. Each day there are 200 minutes of threading machine time, 120 minutes of slotting machine time and 150 minutes of craftsman time available. The profit on each type A product is $5 and on each type B product is $12. To maximize profit, please: 1. Formulate the appropriate linear programming problem. 2. Use graphical solution to solve it, please mark corner points clearly. 3. You can draw the graph by hand, take a picture of your solution

Transcript

00:01 To begin this here, we want to come up with a number of equations to represent the situation.

00:06 There appears to be two major things we're talking about.

00:10 Type a and type b, and each of them gives you time with a threading machine, time with the slotting machine, and time with the craftsman time.

00:18 So for type a, what we have here is the 120 or the 20 minutes for threading machine.

00:30 Okay.

00:30 Then we have 10 minutes on the slotting machine.

00:37 And then finally we have the 7 .5 minutes for craftsmen.

00:42 Okay.

00:43 And i'm assigning them a letter just to keep in mind.

00:46 All right.

00:46 And then in type b, we have something very similar.

00:50 In this case, it was the 10 minutes on threading.

00:55 Then we have 20 minutes for shredding or slotting and then 30 minutes for craftsmen.

01:05 So what we do then is put them together for our next piece, which says each day, there are 200 minutes of threading machine time.

01:14 So let's let x represent type a and we'll let y represent type b.

01:21 So for the first piece, which is the threading machine, it has a maximum of 200 minutes.

01:27 That means the 20 minutes per type a plus the 10 minutes per type b has to be less than equal to the 200 minutes.

01:38 Then we're going to repeat that process for the other two.

01:41 So 120 minutes for slotting.

01:43 So that would be 10 for type a plus 20 for type b.

01:47 That has to be less than or equal to 120.

01:51 Then from there it says we have 150 a craftsman.

01:54 So we would have 7 .5 for type a plus 30 for type b has to be less than or equal to the 150.

02:05 So the last thing we need to know here is that the amount.

02:10 Of type a and the amount of type b can't be negative.

02:12 So we'd have x is greater than equal to zero.

02:15 And y has to be greater than equal to zero.

02:16 So we're focusing really in quadrant one.

02:19 And then from there, it says that the profit of each type a product is five.

02:23 And for type b is 12.

02:26 So that comes up with our equation that we're going to find our maximum in for.

02:31 So it would be z is equal to.

02:33 And again, it would be five for type a.

02:36 And then plus 12 for type b.

02:39 So 12 y.

02:41 So from there, we're going to put both of these into slope intercept form.

02:45 So for the very first one, we're going to subtract 20 to the other side and then divide by 10.

02:51 So what i'll have is y is less than or equal to, 20 divided by 10 is 2x, and then 200 divided by 10 is 20.

02:59 So we'll have plus 20.

03:01 Oh, that's a negative there.

03:03 For the second one, again, we'll divide by, subtract 10x and divide by 20.

03:07 So for this one, we'll have y is less than or equal to 10 divided by 20 is a half.

03:11 So i would have negative 1 over 2x.

03:14 And then 120 divided by 20 would get us 6.

03:18 So we would have plus 6 here.

03:21 And finally, that last one, i'm going to subtract set by 7 .5 and divide by 30.

03:26 So 7 .5 out of 30 is actually reduces 1 fourth.

03:30 So it's negative 1 over 4x.

03:32 And then 150 divided by 30 is going to give you 5.

03:37 So now we're going to take a look at this graph over here on the right.

03:41 And we're going to sketch a graph with all these restraints in them.

03:45 So all of them are less than or equal to, so they're all going to shade below.

03:48 And then we'll look for the corners.

03:50 So for the very first one, we're up at 20.

03:52 Then these two are at six and at five.

03:55 So i'm going to go ahead and start with the smallest one here at five.

03:57 So we're going to go up, one, two, three, four, five, get my y intercept.

04:01 And i'm going to go down one to the right, one, two, three, four.

04:05 And down one to the right, one, two, three, four.

04:08 Down one to the right, one, two, three, four.

04:11 So what i think i'm going to do is get a better graph for us to use here.

04:18 There we go.

04:19 So the reason for is we want to make sure we first we get to 20.

04:22 We only want to be in quadrant one, so it's going to be a lot better this way.

04:25 All right.

04:26 So since we were at that point there, so we're at five, so we're going to go down one over right, right, one, two, three, four, and down one to the right, one, two, three, four.

04:37 Down one to the right, one, two, three, four.

04:39 And then down one to the right, one, two, three, four.

04:41 Four one more time one and over four so this is one of our lines and we'll be shading below it all right then the second one i'm going to go right above it at six this one's at negative one half so down one to the right one two down one to the right one two down one to the right down to the right down to the right so this is our second one again it's a solid line all right the last one's all the way up here at 20.

05:09 So this is going to go down to and to the right one...

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