Can someone help me with this simple problem?? (College Algebra)?

To me, this seems like a simple problem to deal with to me just dividing and seeing how much and item can be made! But I keep getting this problem wrong for some reason! Here is the problem:

A clothing manufacturer makes coats, shirts, and slacks. The times required for cutting, sewing, and packaging each item are shown in the table. (tables in the image)

How many of each should be made to use all available labor hours?

coats?Shirts?Slacks?

Attachment image

2 Answers

Relevance
  • sepia
    Lv 7
    1 month ago

    A clothing manufacturer makes coats, shirts, and slacks.

    The times required for cutting, sewing, and packaging each item

    are shown in the table. (tables in the image)

    Cutting (coats shirts slacks) (20 15 10)

    Sewing (coats shirts slacks) (40 20 24)

    Packaging (coats shirts slacks) (5 12 6)

     Time Available (Cutting 137 hr Sewing 238 hr Packaging 75 hr)

    How many of each should be made to use all available labor hours?

    How many of each should be made to use all available labor hours?

    coats?

    Shirts?

    Slacks?

    • Login to reply the answers
  • 1 month ago

    This is a "system of equations" (aka "simultaneous equations") problem--so it's more than just simple division.

    Let x be the number of coats, y be the number of shirts and z the number of slacks manufactured.  Then the cutting times must add up to 137 hr:

    (20/60) x + (15/60) y + (10/60) z = 137

    I rewrote 20 min as 20/60 hours to get all times in the same units, and did the same for 15 and 10 minute times.  That simplifies to:

    (1/3) x + (1/4) y + (1/6) z = 137

    Following the same pattern, the sewing must add up to 338 hr:

    x + (1/2) y + (2/5) z = 338

    ...and the packaging time must add up to 75 hr:

    (1/12) x + (1/5) y + (1/10) z = 75

    Those are the three equations in your linear system.  Use whatever methods you have learned (elimination, substitution, Cramer's Rule) to solve them.

    • Login to reply the answers
Still have questions? Get your answers by asking now.