document.write( "Question 1169679: A manufacturer produces two models of racing bike, B and C, each of which must be processed through two machine shops. Machine shop 1 is available for 120 hours per month and machine shop 2 for 180 hours per month. The manufacture of each bike of type B takes 6 hours in shop 1 and 3 hours in shop 2. The corresponding times for C are 4 and 10 hours, respectively. If the profit is $180 and $220 per bike of type B and C respectively, how should the manufacturer arrange production to maximize total profit? \n" ); document.write( "
Algebra.Com's Answer #794479 by Theo(13342)\"\" \"About 
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objective function is profit, which you want to maximize.
\n" ); document.write( "the equation for that is p = 180 * x + 220 * y
\n" ); document.write( "x represents the number of type B bikes.
\n" ); document.write( "y represents the number of type C bikes.
\n" ); document.write( "machine shop 1 is available for 120 hours and machine shop 2 is available for 180 hours per month.
\n" ); document.write( "manufacture of type B bike takes 6 hours in shop 1 and 3 hours in shop 2.
\n" ); document.write( "manufacture of type C bike takes 4 hours in shop 1 and 10 hours in shop 2.
\n" ); document.write( "this means shop 1 requires 6 hours for type B bike and 4 hours for type C bike and shop 2 requires 3 hours for type B bike and 10 hours for type C bike.
\n" ); document.write( "this leads to two constraint equation.
\n" ); document.write( "they are:
\n" ); document.write( "6x + 4y <= 120 for shop 1
\n" ); document.write( "3x + 10y <= 180 hours for shop 2.
\n" ); document.write( "this can be seen easier in the following table.
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document.write( "bike           type B           type C           capacity\r\n" );
document.write( "type B           6                4              <= 120 hours     \r\n" );
document.write( "type C           3                10             <= 180 hours\r\n" );
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\n" ); document.write( "x and y must both be greater than or equal to 0.
\n" ); document.write( "this leads to the following summary of requirements.
\n" ); document.write( "6x + 4y <= 120
\n" ); document.write( "3x + 10y <= 180
\n" ); document.write( "x >= 0
\n" ); document.write( "y >= 0
\n" ); document.write( "p = 180x + 220y is the objective function.\r
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\n" ); document.write( "\n" ); document.write( "in the desmos.com calculator, .....
\n" ); document.write( "graph the opposite of the inequalities.
\n" ); document.write( "find the region of feasibility.
\n" ); document.write( "that is the unshaded region on the graph.
\n" ); document.write( "evaluate the objective function at each of the corner points.
\n" ); document.write( "the corner point with the greatest value is the solution.\r
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\n" ); document.write( "\n" ); document.write( "the graph is shown below:\r
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\n" ); document.write( "\n" ); document.write( "each point is in (x,y) format.
\n" ); document.write( "at (0,18), the profit is 0*180 + 18*220 = 3960
\n" ); document.write( "at (10,15), the profit is 10*180 + 15*220 = 5100
\n" ); document.write( "at (20,0), the profit is 20*180 + 0*220 = 3600\r
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\n" ); document.write( "\n" ); document.write( "the maximum profit is at (10,15)\r
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\n" ); document.write( "\n" ); document.write( "all the constraints must be met at the point (10,15)
\n" ); document.write( "6x + 4y <= 120 becomes 6*10 + 4*15 <= 120 which becomes 120 <= 120, so this constraint is met.
\n" ); document.write( "3x + 10y <= 180 becomes 3*10 + 10*15 <= 180 which becomes 180 <= 180, so this constraint is met.
\n" ); document.write( "x >= 0 becomes 10 >= 0, so this constraint is met.
\n" ); document.write( "y >= 0 becomes 15 >= 0, so this constraint is met.\r
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\n" ); document.write( "\n" ); document.write( "all the constraints are met, so manufacture of 10 type B bikes and 15 type C bikes is the maximum profit solution.
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