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Question 120805: A small firm produces both AM and AM/FM car radios. The AM takes 15h to produce, and the AM/FM radios take 20h. The number of production hours is limited to 300h 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.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! This same problem came up a few weeks ago, perhaps it will help you.
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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 is the number of AM/FM radios.
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Let x = number of AM radios; let y = number of AM/FM radios
The production hour constraint:
15x + 20y =< 300
Arrange in general form to plot the graph
20y =< 300 - 15x
y =< 300/20 - (15/20)x
y =< 15 - .75x, use this to plot the production hr graph (Purple line)
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Plant's capacity constraint:
x + y =< 18
y =< 18 - x, plot this graph also (Green line)
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Min AM radio production constraint:
x => 4
This will be a vertical line going thru the x axis at +4
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Min AM/FM radio production constraint
y => 3
This will be a horizontal line going thru the y axis at +3, (Dark blue line)
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Here is the graph as presented by these equations except for x => 4,
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The feasibility region:
1. At or below the purple or green line, whichever is lowest
2. At or above the horizontal line
3. At or to the right of the vertical line
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