SOLUTION: A small firm produces both am and am/fm car radios. The am radios take 15 hours to produce, and the am/fm radios take 20 hours to produce. The number of production hours is limite

Algebra ->  Inequalities -> SOLUTION: A small firm produces both am and am/fm car radios. The am radios take 15 hours to produce, and the am/fm radios take 20 hours to produce. The number of production hours is limite      Log On


   

Question 41919: A small firm produces both am and am/fm car radios. The am radios take 15 hours to produce, and the am/fm radios take 20 hours to produce.
The number of production hours is limited to a total of 300 hours per week. The plant's capacity is limited to a total of 18 radios per week, and existing orders require that at least 4 am radios and at least 3 am/fm radios be produced per week.
Write a system of inequalities representing this situation.
Then draw a graph of the feasible region given these conditions, in which x is the number of am radios and y is the number of am/fm radios.
Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
A small firm produces both am and am/fm car radios. 

The am radios take 15 hours to produce, and the am/fm 
radios take 20 hours to produce.

The number of production hours is limited to a total 
of 300 hours per week.

The plant's capacity is limited to a total of 18 
radios per week, 

and existing orders require that at least 4 am radios
and at least 3 am/fm radios be produced per week.

Write a system of inequalities representing this 
situation.

Then draw a graph of the feasible region given these 
conditions, in which x is the number of am radios and 
y is the number of am/fm radios.

>>..The number of production hours is limited to a total of 300
hours per week..<<

 Hours to make x AM radios + Hours to make y AM/FM radios < 300

                                                15x + 20y < 300  

>>..The plant's capacity is limited to a total of 18 radios per 
week..<<  

                             x AM radios + y AM/FM radios < 18
                                                     
                                                    x + y < 18

                                                
>>..at least 4 am radios..<<
                                                        x > 4


                                                          
>>..at least 3 am/fm radios..<<
                                                        y > 3

Graph the four boundary lines, formed by replacing the 
inequality signs with equal signs:

       15x + 20y = 300

           x + y = 18

               x = 4

               y = 3




The feasible region is the region which is

1. above the horizontal line
2. to the right of the vertical line
3. below both the slanted lines

You'll have to shade it yourself.  I can't on here.

Edwin McCravy
AnlytcPhil@aol.com