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
