document.write( "Question 1151913: Pulse Fuels produces two types of additive blends, A and B. Blend A sells at PhP 1200/liter while Blend B sells at PhP 2000/liter. Its manager would like to maximize revenue by producing the right
\n" ); document.write( "combination of the two additive blends.\r
\n" ); document.write( "\n" ); document.write( "Currently, the company outsources its labor due to the seasonal nature of its products. Its contract with its labor provider requires that a minimum of 5,400 hours be utilized every month. If not utilized, the company will still have to pay the equivalent of 5,400 hours. Each liter of Blend A requires 3 hours of labor while each liter of Blend B requires 1.8 hours. On the other hand, its equipment availability is limited to 9,600 hours per month. Each liter of Blend A requires 2 hours of equipment time while each liter of Blend B requires 3 hours of equipment time. Its Marketing Department has estimated that at least 1000 liters per month of Additive Blend A can be sold while the monthly market demand for Additive Blend B is at least 400 liters. Its Operations Manager has informed the GM that the existing configuration of the operating plant (with a 4000-liter/month capacity) will require that one liter of Additive Blend A is produced for every 2 liters of Additive Blend B. Furthermore, the 4000-liter/month plant must operate, at the least, at 67.5% of its monthly capacity to make it operationally viable. \n" ); document.write( "

Algebra.Com's Answer #773879 by ikleyn(52946) $\"\"$ $\"About$

You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( " First, I will write / (copy) all constraints from the post by @Theo.\r
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\n" ); document.write( "\n" ); document.write( " Then I will show that the problem can be solved analytically in a very simple manner and without using any tools from the outside.\r
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document.write( "x = liters of A;  y = liters of B.\r\n" );
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document.write( "The objective function is r = 1200x + 2000y.\r\n" );
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document.write( "r stands for revenue.\r\n" );
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document.write( "This is what you want to maximize.\r\n" );
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document.write( "The constraint functions are:\r\n" );
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document.write( "3x + 1.8y >= 5400.                         (1)\r\n" );
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document.write( "2x + 3y <= 9600.                           (2)\r\n" );
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document.write( "x >= 1000                                  (3)\r\n" );
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document.write( "x >= 400                                   (4)\r\n" );
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document.write( "x + y <= 4000                              (5)\r\n" );
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document.write( "x + y >= 2700 (equals .675 * 4000)         (6)\r\n" );
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document.write( "y = 2x                                     (7)\r\n" );
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document.write( "x >= 0,  y >= 0                            (8)\r\n" );
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document.write( "I start from constraint (1).  I substitute there y = 2x  from (7) to get\r\n" );
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document.write( "    3x + 1.8y >= 5400  --->  3x + 1.8*(2x) >= 5400  --->  6.6x >= 5400  --->  x >=  = 818.18       (9)\r\n" );
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document.write( "Similarly, I treat constraint (2).\r\n" );
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document.write( "    2x + 3y <= 9600  --->  2x + 3*(2x) <= 9600  --->  8x <= 9600  --->  x <= 1200      (10)\r\n" );
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document.write( "Similarly, I treat constraint (5).\r\n" );
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document.write( "    x + y <= 4000  --->  x + 2x <= 4000  --->  3x <= 4000  --->  x <=  = 1333.33       (11)\r\n" );
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document.write( "Similarly, I treat constraint (6).\r\n" );
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document.write( "    x + y >= 2700  --->  x + 2x >= 2700  --->  3x >= 2700  --->  x >=  = 900       (12)\r\n" );
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document.write( "Thus from (3), (9) - (12)  I have finally  1000 <= x <= 1200.      (13)\r\n" );
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document.write( "With the constraint (7), the revenue function is  r = 1200x + 2000y = 1200x + 2000*(2x) = 1200x + 4000x = 5200x.\r\n" );
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document.write( "So, I want to find maximum value of the revenue function  r = 5200x  at the interval  1000 <= x <= 1200.\r\n" );
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document.write( "It is absolutely clear that the maximum is at the endpoint x= 1200  and the maximum value is then r = 5200*1200 = 6240000.\r\n" );
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document.write( "ANSWER.  The optimum solution is  x= 1200 liters brand A;  y= 2x = 2400 liters of brand B  and the maximum revenue is 6240000 PhP.\r\n" );
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\n" ); document.write( "\n" ); document.write( "The solution is completed, with all imposed constraints (!)\r
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document.write( "    The constraint (7) reduces the problem from 2D (two variable on the plane) to 1D (one variable ONLY (!) )\r\n" );
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