document.write( "Question 1175634: A manufacturer makes two items,A and B. Item A requires 3 minutes of labor to assemble and B requires 4 minutes of assembly time. Item A costs $2 in raw materials and B costs $1. There is a maximum of 3,000 labor minutes available for assembly and a budget of $1,000 in raw material costs per day. Assuming they sell all the produce and that the profit is $5 per item A and $4 per item B,how many of each item must be produced in order to maximize profit? \n" ); document.write( "
Algebra.Com's Answer #801286 by ikleyn(52864) You can put this solution on YOUR website! . \n" ); document.write( "A manufacturer makes two items, A and B. Item A requires 3 minutes of labor to assemble and B requires 4 minutes of assembly time. \n" ); document.write( "Item A costs $2 in raw materials and B costs $1. There is a maximum of 3,000 labor minutes available for assembly and a budget \n" ); document.write( "of $1,000 in raw material costs per day. Assuming they sell all the produce and that the profit is $5 per item A and $4 per item B, \n" ); document.write( "how many of each item must be produced in order to maximize profit? \n" ); document.write( "~~~~~~~~~~~~~~~\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( "Let X = # items A; Y = # items B.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "From the condition, we have this formulation of maximization problem:\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( " (1) the objective function to maximize is the profit P = 5X + 4Y dollars.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Restrictions\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( " (2) 3X + 4Y <= 3000 minutes (assembly time)\r\n" ); document.write( "\r\n" ); document.write( " (3) 2X + Y <= 1000 dollars (material cost)\r\n" ); document.write( "\r\n" ); document.write( " (4) X >= 0, Y >= 0.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "You can make a plot of the feasibility domain.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "It is a quadrilateral in QI with the vertices (X,Y) = (0,0), (500,0), (200,600), (0,750).\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "The solution is one of these 4 points, where the objective function (profit) has a maximum.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "You calculate the values of the function P(X,Y) at listed points\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( " P(0,0) = 0\r\n" ); document.write( "\r\n" ); document.write( " P(500,0) = 5*500 + 4*0 = 2500\r\n" ); document.write( "\r\n" ); document.write( " P(200,600) = 5*200 + 4*600 = 3400\r\n" ); document.write( "\r\n" ); document.write( " P(0,750) = 5*0 + 4*750 = 3000.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Comparing these values, you find the optimal point.\r\n" ); document.write( "\r\n" ); document.write( "It is (X,Y) = (200,600), 200 items A and 600 items B, providing maximum profit of 3400 dollars.\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "Solved.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "--------------\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "In this site, there is a lesson\r \n" ); document.write( "\n" ); document.write( " - Solving minimax problems by the Linear Programming method \r \n" ); document.write( "\n" ); document.write( "which explains, for beginners, metodology of solving such problems in more details.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |