SOLUTION: Caleb runs a factory that makes Blu-ray players. Each R80 takes 8 ounces of plastic and 2 ounces of metal. Each FS20 requires 4 ounces of plastic and 4 ounces of metal. The factory

Algebra ->  Probability-and-statistics -> SOLUTION: Caleb runs a factory that makes Blu-ray players. Each R80 takes 8 ounces of plastic and 2 ounces of metal. Each FS20 requires 4 ounces of plastic and 4 ounces of metal. The factory      Log On


   



Question 1193899: Caleb runs a factory that makes Blu-ray players. Each R80 takes 8 ounces of plastic and 2 ounces of metal. Each FS20 requires 4 ounces of plastic and 4 ounces of metal. The factory has 320 ounces of plastic, 272 ounces of metal available, with a maximum of 16 R80 that can be built each week. If each R80 generates $4 in profit, and each FS20 generates $5, how many of each of the Blu-ray players should Caleb have the factory make each week to make the most profit?
R80:
FS20:
Best profit:

Answer by ikleyn(52874) About Me  (Show Source):
You can put this solution on YOUR website!
.
Caleb runs a factory that makes Blu-ray players.
Each R80 takes 8 ounces of plastic and 2 ounces of metal.
Each FS20 requires 4 ounces of plastic and 4 ounces of metal.
The factory has 320 ounces of plastic, 272 ounces of metal available,
with a maximum of 16 R80 that can be built each week.
If each R80 generates $4 in profit, and each FS20 generates $5,
how many of each of the Blu-ray players should Caleb have the factory make each week
to make the most profit?
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Let x be the number of R80;  let y be the number of FS20.


Restrictions:


    8x + 4y <= 320        (ounces of plastic restriction)

    2x + 4y <= 272        (ounces of metal restriction)

    x <= 16               (with a maximum of 16 R80 that can be built each week)


Profit function:: P(x,y) = 4x + 5y


Rearrange the restrictions

    y <= -2x     + 80    
    y <= (-1/2)x + 68    
    x <= 16


Graph in the first quadrant with horizontal axis x and vertical axis y


    



The feasibility domain is the pentagon in Quadrant I bounded by the red line, green line and blue line.


It has five vertices

     P1 = (0,68)      (y-intercept of green line)

     P2 = (8,64)      (intersection of green line and red line)

     P3 = (16,48)     (intersection of red line and blue line)

     P4 = (16,0)

     P5 = (0,0)


Now calculate the profit function at these points.

The common sense says that it is enough to check the profit function values at points P1, P2, P3 and P4:


    at P1:  P(0,68)  = 4*0  + 5*68 = 340;

    at P2:  P(8,64)  = 4*8  + 5*64 = 352;

    at P3:  P(16,48) = 4*16 + 5*48 = 304.

    at P4:  P(16,0)  = 4*16 + 5*0  =  64.


Thus, the maximum profit is achieved at P2.


Therefore,  x = 8,  y = 64  is the optimal solution.  It gives the profit of 352 dollars.


ANSWER.  8 R80,  64 FS20;  best profit $352.

Solved.

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To see many other similar problems, solved by the same method, look into the lesson
    - Solving minimax problems by the Linear Programming method
in this site.