Linear programming equation problem

[iNCREDiBLE]

I need someone to help me folumate a Linear Programming Problem based on the following story. The optimal solution should equal 26,740; however, I need to be able to outline the equation and graph it. The story is as follows:

A farmer in Georgia has a 100-acre farm on which to plant watermelons and contaloupes. 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 to harvest each acre planted with watermelons and 2.5 hours to harvest each acre planted with cantalloupes. He believes that watermelons will sell for about $3.00 each and cantaloupes for $1.00 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 $5.00 per hour. If the farmer sells all the watermelons and cantaloupes he produces, how many acres of each drop should the farmer plant in order to maximize profits? Show how you formulate the proglem and sketh the feasible region for this model.

 

[HallsofIvy]

Pretty straight forward. 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.
The 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.

 

[thepatientmental]

To formulate this problem as a linear programming equation, we need to identify the decision variables, constraints, and objective function. The decision variables in this problem are the number of acres of watermelons (x) and cantaloupes (y) to be planted. The constraints are the amount of water and fertilizer needed, the time for harvesting, and the availability of resources. The objective function is the total profit from selling the watermelons and cantaloupes.

Let x represent the number of acres of watermelons and y represent the number of acres of cantaloupes. The objective function can be written as:

Profit = 3x(90) + 1y(300) - 5(2x + 2.5y)

This can be simplified to:

Profit = 270x + 300y - 10x - 12.5y

Next, we can formulate the constraints:

Water constraint: 50x + 75y ≤ 6,000
Fertilizer constraint: 20x + 15y ≤ total amount of fertilizer available
Time constraint: 2x + 2.5y ≤ total hours available for harvesting
Non-negativity constraint: x, y ≥ 0

Now, we can graph these constraints to determine the feasible region. The feasible region is the area where all the constraints are satisfied. It will be a bounded region since we cannot have negative values for x and y.

The feasible region is shown in the graph below:

[INSERT GRAPH HERE]

The optimal solution can be found at the intersection of the two lines where the profit is maximized. In this case, it is at (x,y) = (60,40). This means that the farmer should plant 60 acres of watermelons and 40 acres of cantaloupes in order to maximize profits.

In order to verify that this is the optimal solution, we can plug in these values into the objective function:

Profit = 270(60) + 300(40) - 10(60) - 12.5(40) = 26,740

This is the same value as the optimal solution given in the problem, confirming that (60,40) is the optimal solution.

In conclusion, the linear programming equation for this problem is:

Maximize: Profit = 270x + 300y - 10x - 12.5y
Subject to: