Comments on: A small firm produces both AM and AM/FM car radios. The AM radios take 15 h? http://www.regeneratedradio.com/radios-fm/a-small-firm-produces-both-am-and-amfm-car-radios-the-am-radios-take-15-h Tue, 07 Feb 2012 17:03:49 -0500 http://wordpress.org/?v=2.8.4 hourly 1 By: Stop S http://www.regeneratedradio.com/radios-fm/a-small-firm-produces-both-am-and-amfm-car-radios-the-am-radios-take-15-h/comment-page-1#comment-2753 Stop S Tue, 23 Feb 2010 06:09:59 +0000 http://www.regeneratedradio.com/radios-fm/a-small-firm-produces-both-am-and-amfm-car-radios-the-am-radios-take-15-h#comment-2753 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.<br><b>References : </b><br> 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.
References :

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