A farmer in Georgia has a 100-acre farm on which to plant watermelons and cantal

Question

A farmer in Georgia has a 100-acre farm on which to plant watermelons and

cantaloupes. Every acre planted with watermelons requires 50 gallons of water per day

and must be prepared for planting with 20 pounds of fertilizer. Every acre planted with

cantaloupes requires 75 gallons of water per day and must be prepared for planting with

15 pounds of fertilizer. The farmer estimates that it will take 2 hours of labor to harvest

each acre planted with watermelons and 2.5 hours to harvest each acre planted with

cantaloupes. He believes that watermelons will sell for about $3 each, and cantaloupes

will sell for about $1 each. Every acre planted with watermelons is expected to yield

90 salable units. Every acre planted with cantaloupes is expected to yield 300 salable

units. The farmer can pump about 6,000 gallons of water per day for irrigation purposes

from a shallow well. He can buy as much fertilizer as he needs at a cost of $10 per 50-

pound bag. Finally, the farmer can hire laborers to harvest the fields at a rate of $5 per

hour. The farmer sells all the watermelons and cantaloupes he produces and would like

to determine how many acres of each crop to plant in order to maximize profits.

a) Formulate an LP model for this problem.

b) Implement the model in an Excel spreadsheet. What is the optimal solution?

c) Consider the following prices for the watermelons: (i) $2.75/unit, (ii) $2.50/unit,

(iii) $2.25/unit, (iv) $2.00/unit. Which price points (if any) would result in the

farmer maintain his planting strategy? Which price points (if any) should the

farmer respond by changing his planting strategy? Please explain your reasoning.

d) Suppose the farmer can lease up to 20 acres of land from a neighboring farm to

plant additional crops. How many acres should the farmer lease and what is the

maximum amount he should pay to lease each acre?

Explanation / Answer

Answer:

Let C be the number of acres in cantaloupes and W the number of acres in watermelons. Obviously, since he has only 100 acres of land,we must have

C+ W<= 100.

Now look at the water:

Every acre planted with watermelons requires 50 gallons of water per day so W acres requires 50W gallons. Every acre planted with cantaloupes requires 75 gallons of water per day so C acres requires 75C gallons for a total of 50W+ 75C gallons.

The farmer can pump about 6,000 gallons of water per day

so 50W+ 75C<= 6000.

There is no limit on how much fertilizer or how many laborers he can hire so those affect profit but are not restraints.

To find the "feasible region", graph the lines for W+ C= 100 and 50W+ 75C= 6000. Since C and W can't be negative, the lines C= 0 and W= 0 are the other two boundaries. Every acre planted with watermelons is expected to yield 90 salable units and He believes that watermelons will sell for about $3.00 each:

Each acre in watermelon will produce (90)(3)= $270 each so W acres will bring in $270W. Every acre planted with watermelons must be prepared for planting with 20 pounds of fertilizer and He can buy fertilizer at a cost of $10 per 50-pound bag.

Okay, the total amount of fertilizer for W acres would be 20W which is

(20/50)W 50-pound bags which cost (10)(20/50)W= $4W.

he farmer estimates that it will take 2 hours to harvest each acre planted with watermelons and the farmer can hire laborers to harvest the fields at a rate of $5.00 per hour.

W acres will require 2W hours at a cost of (5)(2W)= $10W.

The net profit on W acres of watermelons will be 270W-4W-10W= $256W.

Every acre planted with cantaloupes is expected to yield 300 salable units and He believes that cantaloupes will sell for $1.00 each:

Each acre in cantaloupes will produce (300)(1)= $300 each so C acres will bring in $300C. Every acre planted with cantaloupes must be prepared for planting with 15 pounds of fertilizer and He can buy fertilizer at a cost of $10 per 50-pound bag.

Okay, the amount of fertilizer for C acres would be 15C which is (15/50)C 50-pound bags

which cost (10)(15/50)C= $3C.

The farmer estimates that it will take 2.5 hours to harvest each acre planted with cantalloupesand the farmer can hire laborers to harvest the fields at a rate of $5.00 per hour. C acres will require 2.5C hours at a cost of (5)(2.5C)= $12.50C The net profit for C acres of canteloupes will be 300C- 3C- 12.5C= $284.5C.

The net profit for W acres of watermelons and C acres of canteloupse will be 256W+ 284.5C and that is your object function.

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