Question 113312
A small firm produces both AM and AM/FM car radios. 
The AM radios take 15 h to produce, and the AM/FM radios take 20 h. 
The number of production hours is limited to 300 h 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 the number of AM/FM radios. 
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Domain and Range:
4<=x<=18
3<=y<=18
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Quantity Inequality:
x+y<=18
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Production Inequality: 
15x +  20y <= 300
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Solve Quantity and Production for "y" :
y <= -x+18
y <= (-3/4)x + 15
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Graph:
Draw vertical lines at x = 4 and at x = 18
Draw horizontal lines at y = 3 and at y = 18
Note: These limit the domain and the range of the solution set
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Graph the boudary lines of the Quantity and production Inequalities:
{{{graph(400,300,-10,30,-10,30,-x+18,(-3/4)x+15)}}}
Darken the area in the feasible solution area that is below both
of these boundary lines.
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Cheers,
Stan H.