SOLUTION: a grower has 50 acres of land for which she plans to raise three crops. it costs $200 to produce an acre of carrots and the profit is $60 per acre. it costs $80 to produce an acre

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Question 1174694: a grower has 50 acres of land for which she plans to raise three crops. it costs $200 to produce an acre of carrots and the profit is $60 per acre. it costs $80 to produce an acre of celery and the profit is $20 per acre. finally, it costs $140 to produce an acre of lettuce and the profit is $30 per acre. use the simplex method to find the number of acres of each crop she should plant in order to maximize her profit. assume that her cost cannot exceed $10,000.
Answer by ikleyn(52806) About Me  (Show Source):
You can put this solution on YOUR website!
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a grower has 50 acres of land for which she plans to raise three crops.
it costs $200 to produce an acre of carrots and the profit is $60 per acre.
it costs $80 to produce an acre of celery and the profit is $20 per acre.
finally, it costs $140 to produce an acre of lettuce and the profit is $30 per acre.
use the simplex method to find the number of acres of each crop she should plant in order to maximize her profit.
assume that her cost cannot exceed $10,000.
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x acres for carrot;

y acres for celery;

z acres for lettuce.


The profit function

P(x,y,z) = 60x + 20y + 30z            (1)    (= objective function)



Restrictions:

x   +     y +    z <= 50              (2)   (50 acres in all)

200x +  80y + 140z <= 10000           (3)   (cost to produce)

x >= 0,  y >= 0,  z >=  0             (5)   (standard non-negativity restrictions)



Now go to the site https://www.zweigmedia.com/RealWorld/simplex.html

                   https://www.zweigmedia.com/RealWorld/simplex.html

and use the 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 = 50 acres;  Y = 0 acres;  Z = 0 acres;  p = $3000.    ANSWER


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The input to the solver is shown/presented/documented below : 

Maximize p = 60x + 20y + 30z subject to
x + y + z <= 50 
200x + 80y + 140z <= 10000
x >= 0  
y >= 0
z >= 0


For reliability purposes, I checked this solution by running another solver,

https://www.wolframalpha.com/widgets/view.jsp?id=daa12bbf5e4daec7b363737d6d496120

and obtained the same result.

Solved.

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