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Algebra/Algebra Two: constraints


You are sewing doll clothes to sell at a craft show. Party dresses rake 2.5 hours to make while casual sets take 1 hour. You make a profit of 9 dollars on each party dress and 4 dollars on each casual set. If you have no more than 30 hours available to sew and can make at most 15 outfits to sell, how many of each kind should you sew to maximize your profit?

Hi Brianna, I will be happy to help.

Since you do not know how many party dresses or casual sets to make you must use variables.
Let p be the number of party dresses, and
Let c be the number of casual sets

To find the total time, in hours, it takes to make party dresses, multiply the time it takes to make one dress, 2.5, by the number of dresses, p, to get
Similarity, 1c or c is the total time it takes to make casual sets
All together, you have at most 30 hours to spend on making the outfits, so the sum of the two must be less than or equal to 30, or
2.5p + c <= 30

Let P, capital p, be total profit
In a similar fashion,
P = 9p + 4c

p + c <= 15

Here are your two inequalities
2.5p + c <= 30
p + c <= 15

If it helps, graph both inequalities to visualize the solution.

To solve without graphing:
Notice from the Profit formula, P = 9p + 4c, you make the most from party dresses, so you want the greatest p possible.  Because p and c are real things, outfits, neither p or c can be negative.  

From the first inequality, if you set c = 0, p <= 30/2.5 or p <= 12
Therfore for maximum profit, set p = 12 and c = 0 and plug it into the profit equation to get
P = 108

Please let me know if you have any questions.

I hope this helps



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Anne Losch


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