SOLUTION: A clothing manufacturer has 100 m of silk and 180 m of wool.
To make a suit requires 2m of silk and 3m of wool, and to make a dress requires 1m of silk and 2m of wool. If the prof
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Question 250218: A clothing manufacturer has 100 m of silk and 180 m of wool.
To make a suit requires 2m of silk and 3m of wool, and to make a dress requires 1m of silk and 2m of wool. If the profit on a suit is $108 and the profit on a dress is $60, how many suites and dresses should the manufacturer make to maximize a profit?
My answer
Let P Rep the profit
Let X Rep the number of Suits
Let Y Rep the number of Dresses
Constraints
2x+1y < or = 100
3x +2y < or = 180
x< or = 0
y< or = 0
p=108x + 60y
2x+1y=100 *3
3x+2y=180 *2
_____________
6x+3y=300
6x+4y=360
-------------
-1y= -60
y=60
2x+1(60)=100
2x=40
x=20
the manufacturer should make 20 suits and 60 dresses to maximize profit
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
You got the correct answer, but I think you got there by chance. If you graph the four constraint inequalities, you will see that you have an area of feasibility described by the quadrilateral with vertices (0,0), (0,90), (20,60), and (50,0).
An elementary Linear Optimization theorem says that the optimum values are at a vertex of the feasibility polygon. Because of the particular profit values given, the actual optimum for this problem occurs at the intersection of the two resource constraint boundary lines. But if the profit numbers change, you could have a different result. Completing this problem requires that you check the value of the Profit function for EACH of the feasibility vertices to verify that indeed you have chosen the one that optimizes profit.
As I mentioned, your answer is correct because:
\ +\ 60(60)\ =\ 5760)
\ +\ 60(90)\ =\ 5400)
\ +\ 60(0)\ =\ 5400)
and
\ +\ 60(0)\ =\ 0)
However, if the profit numbers change you can get a different result. Let the profit on suits remain at 108, but the profit on dresses increase to 75. Then:
\ +\ 75(60)\ =\ 6660)
\ +\ 75(90)\ =\ 6750)
\ +\ 75(0)\ =\ 5400)
and
And you would want to manufacture zero suits and 90 dresses.
On the other hand, if the profit on suits were to increase to 130 leaving the profit on dresses at 60, then:
\ +\ 60(60)\ =\ 6200)
\ +\ 60(90)\ =\ 5400)
\ +\ 60(0)\ =\ 6500)
and
So you would want to manufacture 50 suits and zero dresses.
Lesson: Check all your possible solutions before making a judgement.
John
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