A long sleeve blouse requires 1.5 hrs of cutting and 1.2 hrs of sewing. A short sleeve blouse requires 1 hr of cutting and .9 hrs of sewing. Sleeveless blouse requires .5 hrs of cutting and .6 hrs of sewing. There are 380 hrs of labor available in the cutting department each day and 330 hrs in the sewing department. If the plant is to run at full capacity how many of each type of blouse should be made each day.

Let the number of long-sleeve blouses be x

let the number of short-sleeve blouses be y
let the number of sleeveless blouses by z

I see two equations
1.5x + y + .5z = 380
3x + 2y + z = 760 (#1)

1.2x + .9y + .6z = 330 ----> (times 10/3)
4x + 3y + 2z = 1100 (#2)

I doubled the first, then subtracted the second to get

2x + y = 420
or y = 420 - 2x, where 0 ≤ x ≤ 210
and form #1
z = 760 - 3x - 2y , where z > 0

Making a chart of ordered triplets, it is clear that there are many different triplets that are possible for full capacity.

e.g. (150,120,70) , (80,260,0) , (210,0,130) all work.

Are you sure there wasn't a restraint based on the profit of each of the products?

There was nothing. I gave the whole problem. I did some problems earlier and was ok We never studeied this type of problem and I get confused when new shows up.

To determine the number of each type of blouse that should be made each day, we need to set up equations based on the available labor hours in the cutting and sewing departments.

Let's denote the number of long sleeve blouses as L, short sleeve blouses as S, and sleeveless blouses as V.

In the cutting department, a long sleeve blouse requires 1.5 hours, a short sleeve blouse requires 1 hour, and a sleeveless blouse requires 0.5 hours. We have 380 hours available in the cutting department each day. Thus, we can set up the following equation:

1.5L + 1S + 0.5V ≤ 380

In the sewing department, a long sleeve blouse requires 1.2 hours, a short sleeve blouse requires 0.9 hours, and a sleeveless blouse requires 0.6 hours. We have 330 hours available in the sewing department each day. Setting up a similar equation, we get:

1.2L + 0.9S + 0.6V ≤ 330

To find the number of each type of blouse, we need to solve these equations simultaneously.

One approach is to use a method called Linear Programming to find the optimal solution. However, this can be quite complex. Alternatively, we can solve these equations by trial and error.

Let's try different values for L, S, and V and see which combinations satisfy the constraints:

1. Start by assuming all long sleeve blouses: L = 1, S = 0, V = 0
- In the cutting department: 1.5 * 1 + 0 * 1 + 0 * 0.5 = 1.5 hours (less than 380)
- In the sewing department: 1.2 * 1 + 0 * 0.9 + 0 * 0.6 = 1.2 hours (less than 330)
This combination satisfies the constraints.

2. Increase the number of long sleeve blouses and decrease the number of short sleeve blouses:
- L = 2, S = 0, V = 0
- In the cutting department: 1.5 * 2 + 0 * 1 + 0 * 0.5 = 3 hours (less than 380)
- In the sewing department: 1.2 * 2 + 0 * 0.9 + 0 * 0.6 = 2.4 hours (less than 330)
This combination satisfies the constraints.

Continue this trial and error process, increasing the number of long sleeve blouses and decreasing the number of short sleeve blouses, until you find a combination that satisfies all the constraints.

By solving the equations using this method, you will be able to determine the optimal number of each type of blouse that should be made each day to run the plant at full capacity.