Linear Programming #3

Question

A small tailors’ company wants to use at least 130 yards of fabric to sew evening skirts and dresses. A dress requires 4 yards of fabric and the production of a skirt will need 3 yards. Research shows that they will be able to sell at most three times as many skirts as dresses . A dress will take 10 hours to produce and a skirt will take 1 hour. They can assign to this work no more than 286 hours. Each dress will sell for 540,andeachskirtwillsellfor 540 a n d e a c h s k i r t w i l l s e l l f o r 180. How many skirts should they sew to maximize the profit?

Click this - https://www.desmos.com/calculator/0a8b743iyi

It will show the equations I made with it graphed. I need help here. What should I do now?

Also - C(x,y) = 540x+180y (right)?

For the vertices I got - (26,8.667),(27.667,9.226),(0,43.333)and (0,286)

Do I know need to plug these into the C(x,y)...? Once I find the largest, some of the points have fractions. Do i round up???

If you round up, you'll no longer be within the constraints. — The Chaz 2.0, Feb 27 '16 at 16:41
Honestly, I do these every year with my students and can't recall a time when fractions came into play. I would round down, then calculate the C value of each rounded point. — The Chaz 2.0, Feb 27 '16 at 16:43
No. You can remove all the numbers after the decimal places on your own! Then calculate C for each point and compare. — The Chaz 2.0, Feb 27 '16 at 16:47
I got that (27,9) would work but it says 9 is not correct??? — Dan, Feb 27 '16 at 16:57

score 0 · Answer 1 · answered Feb 27 '16 at 16:59

0

Solve the problem getting real numbers for the number of skirts and dresses. Let's say the optimal solution says "35 4/9ths dresses". That's of course not possible. The number of dresses is either ≤ 35 or ≥ 36, not in between.

So you duplicate your problem, once adding the inequality "dresses ≤ 35" and once adding the inequality "dresses ≥ 36". Solve both, and pick the solution that is better. Each solution might have a non-integer value for the number of skirts. In that case you need to do the same again.

answered Feb 27 '16 at 16:59

gnasher729

10,113

What do u mean by solve and pick the one that is better?? could u give me what i should do in with real vertices in the problem? – Dan Feb 27 '16 at 17:04
read what i said above – Dan Feb 27 '16 at 17:10

score 0 · Answer 2 · answered May 27 '19 at 15:20

You implemented the constraint on the number of skirts and dresses incorrectly.

The number of skirts is at most 3 times the number of dresses. So if you can't make 286 skirts and 0 dresses since 286 is not less than 3*0.

Let dresses be x and skirts y, then if you make 20 dresses, you make at most 60 skirts

The correct constraint is y<=3x.

Then your vertices are (10,30), (22,66), and (28,6). You can take it from there.

Linear Programming #3

2 Answers2