document.write( "Question 65561: Draw the appropriate graphs in each of the following.
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document.write( " 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 4AM 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.
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Algebra.Com's Answer #46139 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A small firm produces both AM and AM/FM car radios. \n" ); document.write( "The AM radios take 15 h to produce, and the AM/FM radios take 20 h. \n" ); document.write( "The number of production hours is limited to 300 h per week. \n" ); document.write( "The plant’s capacity is limited to a total of 18 radios per week, and existing orders require that at least 4AM radios and at least 3 AM/FM radios be produced per week. \n" ); document.write( "Write a system of inequalities representing this situation. \n" ); document.write( "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( "----------------- \n" ); document.write( "Production equation: 15x+20y<=300 \n" ); document.write( "Quantity equations: x+y<=18 \n" ); document.write( "AM radio restrictions: 4<=x<=18 \n" ); document.write( "FM radio restrictions: 3<=y<=18 \n" ); document.write( "---------- \n" ); document.write( "Solve where you can for y in terms of x: \n" ); document.write( "y<=(-15/20)x+15 \n" ); document.write( "y<=-x+18 \n" ); document.write( " \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |