Question 1194596: A farmer owns a 100 acre farm and plans to plant at most three crops. The seed for crops A,B, and C costs $40, $20, and $30 per acre, respectively. A maximum of $3200 can be spent on the seed. Crops A,B, and C require 1,2, and 1 work days per acre, respectively, and there are a maximum of 160 workdays available. If the farmer can make a profit of $100 per acre on crop A, $300 per acre on crop B, and $200 per acre on crop
Formulate LPP
Answer by ikleyn(52803)   (Show Source):
You can put this solution on YOUR website! .
A farmer owns a 100 acre farm and plans to plant at most three crops.
The seed for crops A,B, and C costs $40, $20, and $30 per acre, respectively.
A maximum of $3200 can be spent on the seed.
Crops A,B, and C require 1,2, and 1 work days per acre, respectively,
and there are a maximum of 160 workdays available.
If the farmer can make a profit of $100 per acre on crop A, $300 per acre on crop B,
and $200 per acre on crop
Formulate LPP
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Solution
x acres of crop A;
y acres of crop B;
z acres of crop C.
The profit function
P(x,y,z) = 100x + 300y + 200z (1) (= objective function, dollars)
Restrictions:
x + y + z <= 100 (2) (total area, acres)
40x + 20y + 30z <= 3200 (3) (cost of seed, dollars)
1x + 2y + 1z <= 160 (4) (workdays)
x >= 0, y >= 0, z >= 0 (5) (standard non-negativity restrictions)
Now go to the site https://www.zweigmedia.com/RealWorld/simplex.html
and use free of charge solver there.
Input the profit function and the restrictions and press the "Solve" button.
It will solve this maximization problem using the Linear Programming method / (the "simplex method").
The solver produces this solution (this answer)
X = 0 (acres for A); Y = 60 acres for B; Z = 40 acres for C; p = $26000. ANSWER
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The input to the solver is shown/presented/documented below :
Maximize p = 100x + 300y + 200z subject to
x + y + z <= 100
40x + 20y + 30z <= 3200
1x + 2y + 1z <= 160
x >= 0
y >= 0
z >= 0
Solved.
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