document.write( "Question 1198959: A factory makes two types of beds, type A and type B. Each month, a number a number of type A and a number of type B are produced. The following constraints control monthly production:
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document.write( "No more than 50 beds of Type A and no more than 40 beds of type B can be made.
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document.write( "At least 60 beds in all must be made.
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document.write( "The maximum number of beds that can be produced is 80.\r
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document.write( "The profit on type A is Php300 and on type B is Php150. How many beds on both types must be produced to maximize the profit? What is the minimum profit? \n" );
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Algebra.Com's Answer #832647 by math_tutor2020(3817)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Given Facts:
\n" ); document.write( "y = number of beds of type B \n" ); document.write( "where x,y are nonnegative integers\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Fact 1 leads to the inequality \n" ); document.write( "Since x is nonnegative, we can further clarify that \n" ); document.write( "x is in the set {0,1,2,...,49,50}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Fact 2 leads to \n" ); document.write( "y is in the set {0,1,2,...,39,40}\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Fact 3 means \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then fact 4 says \n" ); document.write( "The most beds that can be made is 80.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Fact 5 tells us 300x is the profit for just the type A beds. \n" ); document.write( "Fact 6 tells us 150y is the profit for just the type B beds. \n" ); document.write( "Combine those facts to get 300x+150y as the total profit for both beds.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The goal is to max out P = 300x+150y\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "System of inequalities \n" ); document.write( " \n" ); document.write( "Graph \n" ); document.write( " ![]() \n" ); document.write( "The blue trapezoidal region represents the set of (x,y) points that satisfy all of the inequalities mentioned in the system above. \n" ); document.write( "Points on the boundary are part of the shaded solution set.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The corner points are \n" ); document.write( "A = (20, 40) \n" ); document.write( "B = (40, 40) \n" ); document.write( "C = (50, 30) \n" ); document.write( "D = (50, 10) \n" ); document.write( "Each corner point can be determined using algebra. \n" ); document.write( "For instance, intersect the line y = 40 and x+y = 60 to determine the location of point A(20,40)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "After getting those corner points, we then will plug each into the profit function to see which yields the largest value of P.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If we tried the x and y coordinates of point A, then, \n" ); document.write( "P = 300x+150y \n" ); document.write( "P = 300*20+150*40 \n" ); document.write( "P = 12000 \n" ); document.write( "Repeat for points B through D\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "These are the results you should get
\n" ); document.write( "The min profit happens at point A.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "===============================================\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Answer:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "How many beds on both types must be produced to maximize the profit? 50 of type A and 30 of type B\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "What is the minimum profit? Php 12000\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Edit: just in case you made a typo and are asking for the maximum profit, then that max profit would be Php 19500 \n" ); document.write( " \n" ); document.write( " |