document.write( "Question 71877This question is from textbook
\n" ); document.write( ": A small firm produces both AM and AM/FM car radios. The AM radio take 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 tool 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. \n" ); document.write( "
Algebra.Com's Answer #51409 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
\"15x+%2B+20y+%3C=+300\"
\n" ); document.write( "divide by 5
\n" ); document.write( "\"3x+%2B+4y+%3C=+60\"
\n" ); document.write( "\"x+%3E=+4\"
\n" ); document.write( "\"y+%3E=+3\"
\n" ); document.write( "\"x+%2B+y+%3C=+18\"
\n" ); document.write( "I can determine that
\n" ); document.write( "\"x+%2B+y+%3E=+7\" from above
\n" ); document.write( "I now have 3 inequalities I can plot
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\n" ); document.write( "The feasible area is the smallest area bounded by
\n" ); document.write( "the 3 lines. You can pick values that are just inside
\n" ); document.write( "or just outside to test them by putting into the inequalities
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