Distinguish full capacity from capacity ceilings
A small garment factory manufactures dresses, blouses and T-shirts. The production process includes cutting, sewing and packaging. The table below shows the amounts of time in minutes required to produce each piece of garment.
| Garment | Cutting (min) | Sewing (min) | Packaging (min) |
|---|---|---|---|
| Dress | 45 | 70 | 8 |
| Blouse | 50 | 60 | 6 |
| T-shirt | 20 | 20 | 3 |
The total maximum amounts of time spent on cutting, sewing and packaging are 3350, 4300 and 510 minutes per week respectively.
(i) How many pieces of each garment are produced each week for the factory to operate at full capacity? marks
(ii) Suppose the packaging process is no longer needed and at least 20 pieces of dresses are produced each week. What is the minimum and maximum possible number of T-shirts that can be produced? Assume that the total maximum amounts of time spent on cutting and sewing remains unchanged. marks
[modified 2016/JJC/Promo/1]
Solution
Linear systems
A garment factory makes dresses, blouses and T-shirts. Times in minutes per garment (cutting, sewing, packaging): dress 45, 70, 8; blouse 50, 60, 6; T-shirt 20, 20, 3. The weekly maxima are 3350, 4300 and 510 minutes. How many pieces of each garment are produced each week for the factory to operate at full capacity?
Hint
Let be the numbers of dresses, blouses and T-shirts.
Full capacity means each process uses all its time: three equations.
Solve.
“Full capacity” turns each limit into an equality. With inequalities there would be many solutions.
Answer · (i)
30 dresses,20 blouses,50 T-shirts.
Cutting
Time per dress, blouse and T-shirt times the number of each, equal to the weekly cutting time.
Sewing
Same idea for sewing.
Packaging
And for packaging.
Solve
By calculator. Check cutting: ✓.
Explore the idea
Parametric solutions
Suppose packaging is no longer needed and at least 20 dresses are produced each week. What is the minimum and maximum possible number of T-shirts, if cutting and sewing remain at full capacity?
Hint
Two equations in three unknowns leave a one-parameter family: solve for in terms of .
Impose and .
Read off the range of .
Without the condition that the number of blouses is non-negative you get no upper bound.
The printed solution assumes both cutting and sewing stay at full capacity, but part (ii) only explicitly keeps their maximum times unchanged. Under ceilings, choose 20 dresses and no blouses: and . Cutting limits integer to 122, so the feasible extrema are 0 and 122. At 122, cutting uses 3340 minutes and sewing uses 3840.
Answer · (ii)
10–82 T-shirts at full cutting/sewing capacity; 0–122 if those times are only ceilings.
The source key uses equalities. With capacity ceilings, cutting is binding:900+20z≤3350. Full-capacity whole-number production has z≡2 (mod 8).
Solve in terms of
Subtract the cutting equation from the sewing equation: . Then eliminate between the two equations.
Dresses at least 20
Solve the inequality for .
Blouses cannot be negative
gives .
Conclude
At the factory makes 20 dresses and 45 blouses. At it makes 38 dresses and no blouses. Both are whole numbers. Between these extrema, whole-number production requires ; not every integer between 10 and 82 is feasible under the equalities.
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