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EastWind [94]
4 years ago
5

A carpenter is making tables (x) and chairs (y). Each table takes 5 hours to make and is sold for $100 and each chair takes 3 ho

urs to make and is sold for $30. His goal each week is to make more than $500 from selling his tables and chairs and work no more than 40 hours. Which combination of tables and chairs would maximize his profit?
Question options:

a/ 8 tables and 0 chairs


b/ 2 tables and 10 chairs


c /4 tables and 3 chairs


d/ 5 tables and 0 chairs
Mathematics
2 answers:
sdas [7]4 years ago
8 0
X = tables and y = chairs

5x + 3y < = 40
100x + 30y >  500

well...8 tables and 0 chairs....he will be at 40 hrs and make $ 800....so I believe ur answer is A


AleksandrR [38]4 years ago
6 0

Answer:The solution is region B. The problem can be represented by the equations:

3x + 5y ≤ 40  

5y ≤ -3x + 40

y ≤ -

3

5

x + 8 (graph this inequality)

30x + 100y > 500

100y > -30x + 500

y > -

3

10

x + 5 (graph this inequality)

The shaded regions of the inequality overlap in region B.

Step-by-step explanation:

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