Question 1199320: Bài 7/557) A food producer uses two processing plants, P1 and P2, that
operate 7 days a week. After processing, beef is graded into high-, mediumand low-quality foodstuf s. High-quality beef is sold to butchers, mediumquality beef is used in supermarket ready-meals and the low-quality beef is
used in dog food. The producer has contracted to provide 120 kg of high-,
80 kg of medium- and 240 kg of low-quality beef each week. It costs $4000
per day to run plant P1 and $3200 per day to run plant P2. Each day P1
processes 60 kg of high-quality beef, 20 kg of medium-quality beef and 40kg of low-quality beef. The corresponding quantities for P2 are 20 kg, 20 kg
and 120 kg, respectively. How many days each week should the plants be
operated to fulfi l the beef contract most economically?
Answer by ikleyn(52794)   (Show Source):
You can put this solution on YOUR website! .
A food producer uses two processing plants, P1 and P2, that
operate 7 days a week. After processing, beef is graded into high-, medium and low-quality food stuf s.
High-quality beef is sold to butchers, medium quality beef is used in supermarket, ready-meals and the low-quality beef
is used in dog food.
The producer has contracted to provide 120 kg of high-, 80 kg of medium- and 240 kg of low-quality beef each week.
It costs $4000 per day to run plant P1 and $3200 per day to run plant P2.
Each day P1 processes 60 kg of high-quality beef, 20 kg of medium-quality beef and 40kg of low-quality beef.
The corresponding quantities for P2 are 20 kg, 20 kg and 120 kg, respectively.
How many days each week should the plants be operated to fulfil the beef contract most economically?
~~~~~~~~~~~~~~~~
x = days for plant P1;
y = days for plant P2.
Objective function to minimize is
p(x,y) = 4000x + 3200y dollars.
Constraints
60x + 20y >= 120 (high-quality beef, in kilograms)
20x + 20y >= 80 (medium-quality beef)
40x + 120y >= 240 (low-quality beef)
0 <= x <= 7; 0<= y <= 7 (days to work)
Now go to the site https://www.zweigmedia.com/RealWorld/simplex.html
and use the free of charge solver there.
The input to the solver is shown/presented/documented below :
Minimize p = 4000x + 3200y
60x + 20y >= 120
20x + 20y >= 80
40x + 120y >= 240
x >= 0
x <= 7
y >= 0
y <= 7
Copy-paste it to the solver.
It will solve this minimization problem using the Linear Programming method / (the "simplex method").
The solver produces this solution (this answer)
x = 1 day; y = 3 days; p = $13600. ANSWER
At this point, the solution is complete.
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There are several ways to solve the problem.
For example, since the problem is 2D (for 2 unknown variables), graphics method does work.
On solving using graphics method, see the lesson
- Solving minimax problems by the Linear Programming method
in this site.
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