mrbean1rowan
contestada




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 hours 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

Respuesta :

texaschic101
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

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: