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Question 85530: This is a pretty big problem for me and was wondering if I could get some help with it. I appreciate the help alot. Thank you.
A small firm produces both AM and AM/FM car radios. the AM radios take 15h to produce, and the AM/FM take 20h. The number of production hours is limited to 300 h per week. The plants capacity is limitedto a total of 18 radios per week, and exsisting orders require that at least 4 AM radios and at least 3 AM/FM radios be produced per week. Write a system of inequalities represnting 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.
I know this is a big word problem, and I didn't know where to start. I am hoping I could get some help with this. Thank you again.
Answer by stanbon(75887)   (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 15h to produce, and the AM/FM take 20h. The number of production hours is limited to 300 h per week. The plants capacity is limitedto a total of 18 radios per week, and exsisting orders require that at least 4 AM radios and at least 3 AM/FM radios be produced per week. Write a system of inequalities represnting this situation. Then, draw a graph of the feasible region given these conditions,
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in which x is the number of AM radios and y the number of AM/FM radios.
Production INEQUALITY: x+y<=18
x am radios require 15x hrs to produce
y fm fadios require 20y hrs to produce
Prod Hours INEQUALITY: 15x+20y<=300
AM limit: 4<=x<=18
FM limit: 3<=y<=18
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Draw an x-y coordinate system.
Draw vertical lines at x=4 and x=18
Draw horizontal lines at y=3 and y=18
Solve x+y<=18 for y and graph the line and the halfplane solution set.
Solve 15x+20y<=300 for y and graph the line and the halfplane solution set.
Darken the "feasible" region.
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Cheers,
Stan H.
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