A small firm produces both AM and AM/FM car radios. The AM radios take 15 h?
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.
Let x= the number of AM radios and y= the number of AM/FM radios produced. You need inequality equations for the production hours, the number of radios produced and the minimum number of radios produced.
15x + 20y ≤ 300
x + y ≤ 18
x ≥ 4
y ≥ 3
I would draw the graph but I don’t know how to do that on here. To find the point of where the inequalities cross solve this system of equations:
15x+20y=300
x+y=18
This works out to 12 AM and 6 AM/FM radios per week.
Let x= the number of AM radios and y= the number of AM/FM radios produced. You need inequality equations for the production hours, the number of radios produced and the minimum number of radios produced.
15x + 20y ≤ 300
x + y ≤ 18
x ≥ 4
y ≥ 3
I would draw the graph but I don’t know how to do that on here. To find the point of where the inequalities cross solve this system of equations:
15x+20y=300
x+y=18
This works out to 12 AM and 6 AM/FM radios per week.
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