Algebra Word Problems
posted by mike .
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.