Expert: Paul Klarreich - 2/15/2006

Question
1.  (Answers should be accurate to at least 2 decimal places.) A farmer has a 240 acre farm on which she plants two crops: corn and soybeans. For each acre of corn planted, her expenses are $50 and for each acre of soybeans planted, her expenses are $100. Each acre of corn requires 100 bushels of storage and yields a profit of $60; each acre of soybeans requires 40 bushels of storage and yields a profit of $90.
If the total amount of storage space available is 19200 bushels and the farmer has only $20000 on hand, how many acres of each crop should she plant in order to maximize her profit?  

Answer
Hi, Kevin,
You wrote:

Subject:   
Question:  1. (Answers should be accurate to at least 2 decimal places.) A farmer has a 240 acre farm on which she plants two crops: corn and soybeans. For each acre of corn planted, her expenses are $50 and for each acre of soybeans planted, her expenses are $100. Each acre of corn requires 100 bushels of storage and yields a profit of $60; each acre of soybeans requires 40 bushels of storage and yields a profit of $90.
If the total amount of storage space available is 19200 bushels and the farmer has only $20000 on hand, how many acres of each crop should she plant in order to maximize her profit?

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Notes:  When you send future problems, please include a subject line and state what you are studying (I need a context) and just what you have already tried to do on the problem.
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These are called 'linear programming' programs, and the process goes like this:

1. Select the variables to be 'chosen'.  In this case, since the question asks 'how many ... of each.....' the variables are clearly:

C = number of acres of corn to plant.
S = number acres of soybeans to plant.

2. Identify the quantity to be maximized, in this case the profit, and express it in terms of the variables, taking both costs and revenues into account.  So you do something like this:

One acre of corn planted brings in $60 in profit.
..........  soy ................   $90 ..........

So P = 60C + 90S

3. Identify the constraints -- limits on the acreage because of other considerations:

Total expenses cannot exceed $20000
Total storage  cannot exceet  19200 bushels.
And, of course, total acreage cannot exceed 240.

Write those as inequalities involving the variables, such as:

Expense:  Corn expense = 50C,  Soy expense = 100S, Total = 50C + 100S
Storage:  Corn storage = 100C, soy storage = 40S, Total = 100C + 40S.

Therefore:

50C + 100S <= 20000
100C + 40S <= 19200
and C + S <= 240

4. Do some graphing and solving of simultaneous equations.  When you graph the inequalities, you get a 'feasible region' that looks like a small corner of the first quadrant, since you also have the conditions that:

C >= 0  and S >= 0

In this case it is a pentagon.  Assuming you have C as the horizontal (x-) axis and S as vertical, you have 'corners' at these points:

The origin (0,0)
The C-intercept, where S=0.  (Plant no soy, only corn.)  This will be at C=192.
The S-intercept, where C=0. (No corn.)  This is at S=200.

And, finally, at the points of intersection of the lines whose EQUATIONS are:

50C + 100S = 20000  or   C + 2S = 400
100C + 40S = 19200  or  5C + 2S = 960
C + S = 240

So all you have to do is solve three pairs of simultaneous equations for the  corners:
First:
C + 2S = 400
5C + 2S = 960
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4C      = 560
C = 140

140 + 2S = 400
     2S = 260
      S = 130

So this intersection is at C=140, S=130.  Oops -- that's outside the feasible region. Can't have 270 acres.


Second pair:
C + 2S = 400
C + S  = 240
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    S = 160
    C = 80
That's possible.

Third pair:
5C + 2S = 960
C +  S = 240
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3C      = 480
C = 160
S = 80
Also possible.

Now the final bit of work.  You have these possible points:
1. C = 0,    S = 0
2. C = 192,  S = 0
3. C = 0,    S = 200
4. C = 80,   S = 160
5. C = 160,  S = 80

All you have to do now is take each point and plug it into your expression for P.  (Remember P?  The profit function.)

P = 60C + 90S

Whichever of those five points gives the largest value of P is the desired point and the solution to the problem.  I leave that part to you.  (Hint: It's not the first one.)