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) (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.
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