Question

A California grower has a 50-acre farm on which to plant strawberries and tomatoes. The grower has available 300 hours of labor per week and 800 tons of fertilizer, and he has contracted for shipping space for a maximum of 26 acres’ worth of strawberries and 37 acres’ worth of tomatoes. An acre of strawberries requires 10 hours of labor and 8 tons of fertilizer, whereas an acre of tomatoes requires 3 hours of labor and 20 tons of fertilizer. The profit from an acre of strawberries is $400, and the profit from an acre of tomatoes is $300. The farmer wants to know the number of acres of strawberries and tomatoes to plant. Formulate a linear programming model for this problem and solve it using the graphical method. Your file should contain the following information: LP model where decision variables are clearly identified, a graph showing all constraints with feasible region clearly identified and a table showing all extreme corners in the feasible region with optimal solution.

Answer 1

Let x denotes the no. of acres of strawberries
Let y denotes the no.of acres of tomatoes

In this case, we have to maximize the profit.

The profit function is as follows:

P = 400x + 300y

Based on information provided in the question, following are constraint equations
For Land:
x + y <= 50

For shipping space:
x <= 26
y <= 37

For Fertilizer
8x + 20y <= 800

For Labor:
10x + 3y <= 300

In addition, the no. of strawberries and tomatoes cannot be negative, so:

x >= 0
y >= 0

For graphing all the above equations, they are written in terms of y as follows:
For Land:
y <= 50-x

Equation related to shipping space remains the same at:
x <= 26
y <= 37

For Labor:
3y <= 300-10x becomes:
y <= 100 - 10x/3

For fertilizer
20y <= 800-8x becomes:
y <= 40-8x/20 which becomes:
y <= 40 - 2x/5

Plotting all these constraint equation on graph, we get:

According to LP theory, maximum or minimum points lie at intersection. They never lie in between.

From the graph, we can see that there are 7 intersection points. Following is the table showing profit for each of these intersection points:

x	y	Profit
0	0	0
0	37	11100
7.5	37	14100
16.67	33.33	16667
21.43	28.57	17143
26	13.33	14399
26	0	10400

From the table, we can see that profit will be maximum when:

x = 21.43 and y = 28.57

Farmer should plant strawberries on 21.43 acres

Farmer should plant tomatoes on 28.57 acres

Farmer will earn profit of $17143