Linear Programming Assignment

This article discusses the solution of a linear programming problem using graphical technique. It explains the constraints related to farm land and pesticide usage. It also discusses extreme points and the optimized net profit of the farm. The article is helpful for students studying operations research.

Running Head: LINEAR PROGRAMMING ASSIGNMENT
Linear Programming Assignment
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1LINEAR PROGRAMMING ASSIGNMENT
Answer 7
Part A
The given problem will be solved by using the graphical technique. The following method
described is followed in solving the problem using graphical technique (Mahato 2015). Let x1 be
the number of acres of maize crop and x2 be the number of acres of vegetables that has been
cultivated in the 80 acres of farm land.
The objective function of this problem can be given as follows:
Z=5000 x1 +3000 x2
The total acre of farm land present is 80 acres. Thus, the constraint related to firm land is given
by the following inequality:
x1+ x2 ≤ 80
The total pesticide usage is limited to 100 gallons. Maize plantation requires 2 gallons of
pesticide per acre and vegetable plantation requires 1 gallon of pesticide per acre. Thus, the
pesticide constraint can be given as follows:
2 x1 + x2 ≤ 100
Since, the number of acres of land cannot be negative, the non-negativity constraints are given as
follows:
x1 ≥ 0 , x2 ≥ 0
Part B
Let the equation corresponding to the first constraint be

2LINEAR PROGRAMMING ASSIGNMENT
L1: x1+ x2=80
The origin O (0, 0) satisfies (0 + 0) = 0 < 80, which satisfies the farm land constraint. Thus, the
inequality satisfies all points on L1 and on the origin side of L1.
Let the equation corresponding to the second constraint be
L2: 2 x1 + x2=100
The origin O (0, 0) satisfies (2 * 0 + 0) = 0 < 100, which satisfies the pesticide constraint. Thus,
the inequality satisfies all points on L2 and on the origin side of L2.
At some of the extreme points obtained in the feasible region, the optimal solution will exist.

3LINEAR PROGRAMMING ASSIGNMENT
Extreme Points Co-Ordinates Value of z
A (0, 80) (5000 * 0) + (3000
* 80) = 240000
B (20, 60) (5000 * 20) +
(3000 * 60) =
280000
C (50, 0) (5000 * 50) +
(3000 * 0) =
250000
O (0, 0) (5000 * 0) + (3000
* 0) = 0
The objective function is maximized at B (20, 60). Thus, 20 acres of maize and 60 acres of
vegetables has to be cultivated in order to maximize the profit.
Part C
The optimized net profit of the farm is ($5,000 * 20) + ($3,000 * 60) = 280,000

4LINEAR PROGRAMMING ASSIGNMENT
References
Mahato, D. (2015). Essentials of Operations Research. [online] researchgate.net. Available at:
https://www.researchgate.net/publication/298178045_Linear_Programming_Graphical_Method
[Accessed 25 Jul. 2018].