Question 1118471: Summer runs a factory that makes DVD players. Each S100 takes 6 ounces of plastic and 2 ounces of metal. Each FS20 requires 3 ounces of plastic and 4 ounces of metal. The factory has 234 ounces of plastic, 264 ounces of metal available, with a maximum of 14 S100 that can be built each week. If each S100 generates $13 in profit, and each FS20 generates $13, how many of each of the DVD players should Summer have the factory make each week to make the most profit?
S100:
FS20:
Best profit:
Found 3 solutions by ankor@dixie-net.com, stanbon, ikleyn:
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Summer runs a factory that makes DVD players.
let a = no. of FS20 DVD players
let b = no. of F100's
Each S100 takes 6 ounces of plastic and 2 ounces of metal.
Each FS20 requires 3 ounces of plastic and 4 ounces of metal.
The factory has 234 ounces of plastic, 264 ounces of metal available,
with a maximum of 14 S100 that can be built each week.
Put the equation in the slope/intercept form
The plastic equation
3a + 6b = 234
6b = -3a + 234
b < -.5a + 39
the metal equation
4a + 3b = 264
3b = -4a + 264
b < -1.25a + 88
Max S100 equation
b =< 14
:
Plot these 3 equations
:
:
If each S100 generates $13 in profit, and each FS20 generates $13, how many of each of the DVD players should Summer have the factory make each week to make the most profit?
Three intercepts around the feasibility area
50,14; 60,14; 66,6
:
Profit equations
50(13) + 14(13) = $832
60(13) + 14(13) = $962
66(13) + 6(13) = $936,
:
Max profit when you make 60 FS20 and 14 F100's
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! Summer runs a factory that makes DVD players.
Each S100 takes 6 ounces of plastic and 2 ounces of metal.
Each FS20 rReequires 3 ounces of plastic and 4 ounces of metal.
The factory has 234 ounces of plastic, 264 ounces of metal available, with a maximum of 14 S100 that can be built each week.
---------------------------------
If each S100 generates $13 in profit, and each FS20 generates $13, how many of each of the DVD players should Summer have the factory make each week to make the most profit?
s <= 14
Plastic:: 6s +3f <=234
Metal:::: 2s +4f <= 264
Profit function:: P = 13f + 13s
--------
Rearrange::
f <= -2s + 78
f <= (-1/2)s + 132
------------------------
Graph in the first quadrant with horizontal axis s and vertical axis f
-----
Find the corner coordinates:: (s,f)
Note:: Your Inequalities do not define an enclosed area.
----
Cheers,
Stan H.
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S100:
FS20:
Best profit:
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! Summer runs a factory that makes DVD players.
Each S100 takes 6 ounces of plastic and 2 ounces of metal.
Each FS20 rReequires 3 ounces of plastic and 4 ounces of metal.
The factory has 234 ounces of plastic, 264 ounces of metal available, with a maximum of 14 S100 that can be built each week.
---------------------------------
If each S100 generates $13 in profit, and each FS20 generates $13, how many of each of the DVD players should Summer have
the factory make each week to make the most profit?
~~~~~~~~~~~~~~~~~~~~~
Writing of the other tutor contained a lot of errors, that made the solution IMPOSSIBLE.
I fixed these errors and placed the completed and rectified solution below.
s <= 14
Plastic:: 6s +3f <=234
Metal:::: 2s +4f <= 264
Profit function:: P(s,f) = 13f + 13s
Rearrange::
f <= -2s + 72 <<<---=== here I fixed the other tutor error
f <= (-1/2)s + 66 <<<---=== here I fixed the other tutor another error
Graph in the first quadrant with horizontal axis s and vertical axis f
The feasibility domain is the pentagon in Quadrant I bounded by the red line, green line and blue line.
It has five vertices
P1 = (0,66) (y-intercept of green line)
P2 = (4,64) (intersection of green line and red line)
P3 = (14,44) (intersection of the red line and blue line)
P4 = (14,0)
P5 = (0,0)
Now you need to calculate the profit function at these points.
The common sense says that it is enough to check the profit function values in the points P1, P2 and P3:
at P1: P(0,66) = 0*13 + 66*13 = 858;
at P2: P(4,64) = 4*13 + 64*13 = 884;
at P3: P(14,44) = 13*14 + 13*44 = 754.
Thus, the maximum profit is achieved at P2.
Therefore, s = 4, f = 64 is the optimal solution. It gives the profit of 884 dollars.
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.
* * * This my advise/recommendation is addressed to the tutors of this forum, in the first turn. * * *
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