SOLUTION: A farmer has 1000 acres of land on which he can grow corn, wheat and soybean. Each acre of corn costs birr 100 for preparation, requires 7 man-days work and yields a profit of birr

Algebra ->  Linear-equations -> SOLUTION: A farmer has 1000 acres of land on which he can grow corn, wheat and soybean. Each acre of corn costs birr 100 for preparation, requires 7 man-days work and yields a profit of birr      Log On


   

Question 1151279: A farmer has 1000 acres of land on which he can grow corn, wheat and soybean. Each acre of corn costs birr 100 for preparation, requires 7 man-days work and yields a profit of birr 30. An acre of wheat costs birr 120 to prepare, requires 10 man-days of work and yields a profit of birr 20. An acre of Soybeans costs birr 70 to prepare, requires 8 man-days of work and yields a profit of birr 20. If the farmer has birr 100,000 for preparation and can count on 8000 man-days of work, how may acres should be allocated to each crop to maximize profit?
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
x acres for corn;

y acres for wheat;

z acres for soybeans.


The profit function

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



Restrictions:

x   +     y +   z <= 1000             (2)   (1000 acres in all)

100x + 120y + 70z <= 100000           (3)   (cost for preparation)

7x   +  10y +  8z <= 8000             (4)   (man-days work)

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 solver produces this solution (this answer)


   X = 1000 acres;  Y = 0 acres;  Z = 0 acres;  p = 30000.    ANSWER


--------------

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

Maximize p = 30x + 20y + 20z subject to
x + y + z <= 1000 
100x + 120y + 70z <= 100000
7x + 10y + 8Z <= 8000
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.