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Question 151600: Production Turley Tailor Inc. makes long-sleeve, shortsleeve,
and sleeveless blouses. A long-sleeve blouse
requires 1.5 hours of cutting and 1.2 hours of sewing. A
short-sleeve blouse requires 1 hour of cutting and .9 hour of
sewing. A sleeveless blouse requires .5 hour of cutting
and .6 hour of sewing. There are 380 hours of labor available
in the cutting department each day and 330 hours 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?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! long-sleeve, shortsleeve,
and sleeveless blouses.
:
Let x = l.s blouse
Let y = s.s
Let z = s.l
:
A long-sleeve blouse requires 1.5 hours of cutting and 1.2 hours of sewing. A
short-sleeve blouse requires 1 hour of cutting and .9 hour of
sewing. A sleeveless blouse requires .5 hour of cutting
and .6 hour of sewing. There are 380 hours of labor available
in the cutting department each day and 330 hours 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?
:
The cutting equation:
1.5x + 1y + .5z = 380
:
The sewing equation
1.2x + .9y + .6z = 330
;
We got 3 unknowns and two equation, make an assumption on x:
We want 1.5x and 1.2x values that are multiples of 10;
:
Try x = 100 l.s. blouses
Substitute 100 for x and we have two equations with two unknowns:
150 + 1y + .5z = 380
1y + .5z = 380 - 150
1y + .5z = 230 (1st equation)
and
120 + .9y + .6z = 330
.9y + .6z = 330 - 120
.9y + .6z = 210 (2nd equation)
:
Multiply the 1st equation by 6, and the 2nd equation by 5:
6.0y + 3z = 1380
4.5y + 3z = 1050
-------------------subtraction eliminates z, find y
1.5y + 0z = 330
y = 
y = 220 s.l. blouses
:
Find z using the 1st equation
1(220) + .5z = 230
.5z = 230 - 220
.5z = 10
z = 20 s.l. blouses
;
There may be other solutions, I am not sure about that, but these satisfy:
1.5(100) + 1(220) + .5(20) = 380
150 + 220 + 10 = 380
and
1.2(100) + .9(220) + .6(20) = 330
120 + 198 + 12 = 330
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