Maximizing Profit through Linear Programming in Manufacturing

The task requires maximizing profits in a production scenario by determining the optimal quantities of two products (x1 and x2) subject to resource constraints. The objective function z = 1.27x1 + 1.54x2 is optimized within constraints: 134.4x1 + 188.16x2 ≤ 268.8, 24.96x1 + 8.32x2 ≤ 83.2, and 11.2x1 + 11.2x2 ≤ 56, with x1 ≥ 0 and x2 ≥ 0. The problem is solved graphically to identify feasible regions and using Excel Solver for precise solutions. Additionally, a separate linear programming model involves maximizing z = 4x1 + 10x2 under constraints 6x1 + 5x2 ≤ 120, x1 + 2x2 ≤ 20, and 4x1 ≥ 20. The graphical method identifies the optimal solution at vertex A (5, 7.5) with a maximum profit of $95.

Running Head: PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT
OPERATIONS
Planning Techniques for Logistics and Transport Operations
Name of the Student
Name of the University
Author Note

1PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
Answer 1
Let x1 be the number of jars of Western Foods Salsa produced and x2 be the number of
jars of Mexico City Salsa produced.
The sales revenue earned from each jar of Western Food Salsa is €1.64 and from each jar of
Mexico City Salsa is €1.93.
Each jar of salsa produced weighs 280 gm.
For Western Food Salsa, the blend contains 50% whole tomatoes, 30% tomato sauce and 20%
tomato paste.
Therefore, quantity of whole tomatoes in a jar = 50
100 ×280 gm=140 gm=0.14 kg
Therefore, quantity of tomato sauce in a jar = 30
100 ×280 gm=84 gm=0.084 kg
Therefore, quantity of tomato paste in a jar = 20
100 ×280 gm=56 gm=0.056 kg
The price per kilo of whole tomatoes, tomato sauce and tomato paste are €0.96, €0.64 and €0.56
respectively.
Therefore, cost of whole tomatoes in a jar = € (0.14 × 0.96) = €0.13
Therefore, cost of tomato sauce in a jar = € (0.084 × 0.64) = €0.05
Therefore, cost of tomato paste in a jar = € (0.056 × 0.56) = €0.03
The cost of spices and other ingredients = €0.10
Cost of each empty jar = €0.02

2PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
Cost of labeling and filling each jar = €0.03
Therefore, total cost of each jar of Western Food Salsa = € (0.13 + 0.05 + 0.03 + 0.02 + 0.10 +
0.03) = €0.37
Thus, the profit earned from each jar of Western Food Salsa = Revenue earned – Total Cost
= € (1.64-0.37)
= €1.27.
For Mexico City Salsa, the blend contains 70% whole tomatoes, 10% tomato sauce and 20%
tomato paste.
Therefore, quantity of whole tomatoes in a jar = 70
100 ×280 gm=196 gm=0.196 kg
Therefore, quantity of tomato sauce in a jar = 10
100 ×280 gm=28 gm=0.028 kg
Therefore, quantity of tomato paste in a jar = 20
100 ×280 gm=56 gm=0.056 kg
The price per kilo of whole tomatoes, tomato sauce and tomato paste are €0.96, €0.64 and €0.56
respectively.
Therefore, cost of whole tomatoes in a jar = € (0.196 × 0.96) = €0.19
Therefore, cost of tomato sauce in a jar = € (0.028 × 0.64) = €0.02
Therefore, cost of tomato paste in a jar = € (0.056 × 0.56) = €0.03
The cost of spices and other ingredients = €0.10

3PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
Cost of each empty jar = €0.02
Cost of labeling and filling each jar = €0.03
Therefore, total cost of each jar of Western Food Salsa = € (0.19 + 0.02 + 0.03 + 0.02 + 0.10 +
0.03) = €0.39
Thus, the profit earned from each jar of Western Food Salsa = Revenue earned – Total Cost
= € (1.93-0.39)
= €1.54.
Thus, the profit z in € is given by the following equation:
z = 1.27 x1 + 1.54 x2
Miguel can purchase up to 280 kg of whole tomatoes
Maximum cost for purchasing whole tomatoes = €0.96 × 280 = €268.8
Western food Salsa contains 50% whole tomatoes in each blend and Mexico City Salsa contains
70% in each blend.
Therefore, quantity of whole tomatoes for Western Food Salsa = 50
100 ×280 kg=140 kg
Therefore, cost of whole tomatoes for Western Food Salsa = €0.96 × 140 = €134.4
Therefore, quantity of whole tomatoes for Mexico City Salsa = 70
100 ×280 kg=196 kg
Therefore, cost of whole tomatoes for Mexico City Salsa = €0.96 × 196 = €188.16

4PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
Thus, the inequality for maximum cost of whole tomatoes becomes:
134.4 x1 + 188.16 x2 ≤ 268.8
Miguel can purchase up to 130 kg of tomato sauce
Maximum cost for purchasing tomato sauce = €0.64 × 130 = €83.2
Western food Salsa contains 30% tomato sauce in each blend and Mexico City Salsa contains
10% in each blend.
Therefore, quantity of whole tomatoes for Western Food Salsa = 30
100 ×130 kg=39 kg
Therefore, cost of whole tomatoes for Western Food Salsa = €0.64 × 39 = €24.96
Therefore, quantity of tomato sauce for Mexico City Salsa = 10
100 ×130 kg=13 kg
Therefore, cost of whole tomatoes for Mexico City Salsa = €0.64 × 13 = €8.32
Thus, the inequality for maximum cost of whole tomatoes becomes:
24.96 x1 + 8.32 x2 ≤ 83.2
Miguel can purchase up to 100 kg of tomato paste
Maximum cost for purchasing tomato paste = €0.56 × 100 = €56
Western food Salsa contains 20% tomato sauce in each blend and Mexico City Salsa contains
20% in each blend.

5PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
Therefore, quantity of whole tomatoes for Western Food Salsa = 20
100 ×100 kg=20 kg
Therefore, cost of whole tomatoes for Western Food Salsa = €0.56 × 20 = €11.2
Therefore, quantity of tomato sauce for Mexico City Salsa = 20
100 ×100 kg=20 kg
Therefore, cost of whole tomatoes for Mexico City Salsa = €0.56 × 20 = €11.2
Thus, the inequality for maximum cost of whole tomatoes becomes:
11.2 x1 + 11.2 x2 ≤ 56
Since, the number of jars produced cannot be negative, thus, x1 ≥ 0, x2 ≥ 0.
The profit has to be maximized. Thus, the whole problem can be framed as follows:
Maximize z = 1.27 x1 + 1.54 x2
Subject to the constraints: 134.4 x1 + 188.16 x2 ≤ 268.8
24.96 x1 + 8.32 x2 ≤ 83.2
11.2 x1 + 11.2 x2 ≤ 56
x1 ≥ 0, x2 ≥ 0

6PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
Answer 2
Part a
Let L1: 6 x1 + 5 x2 = 120
Now, the origin, O: (0, 0) satisfies (6 * 0) + (5 * 0) = 0 < 120
Therefore, origin satisfies the inequality 6 x1 + 5 x2 ≤ 120
Hence, 6 x1 + 5 x2 ≤ 120 is satisfied by all points on L1 and on the origin side of L1
Let L2: x1 + 2 x2 = 20
Now, the origin, O: (0, 0) satisfies 0 + (2 * 0) = 0 < 20
Therefore, origin satisfies the inequality x1 + 2 x2 ≤ 20
Hence, x1 + 2 x2 ≤ 20 is satisfied by all points on L2 and on the origin side of L2
Let L3: 4 x1 = 20
Now, the origin, O: (0, 0) satisfies (4 * 0) = 0 < 20
Therefore, origin satisfies the inequality 4 x1 ≤ 20
Hence, 4 x1 ≥ 20 is satisfied by all points on L2 and on the non-origin side of L2
The feasible region is the triangle ABC in the graph.
Therefore, optimal feasible solution exists and is attained at some extreme points of the feasible
region.
Extreme Points Co-Ordinates Value of z

7PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
A (5, 15
2 ) (4 *5) + (10 * 15
2 ) = 95
B (5, 0) (4 * 5) + (10 * 0) = 20
C (20, 0) (4 * 20) + (10 * 0) = 80
The maximum value of z is attained at A.
Therefore, the optimal solution is x1 = 5, x2 = 15
2 and maximum profit = 95.
Figure 1: Graphical solution of the problem
Part b
The given linear programming model can also be solved using solver add-in in Microsoft
excel. A table has been made in excel with the respective headings as given in the figure below.
The formulae that have been used in each of the cells are also shown in the figure. The number
6x1 + 5x2≤ 120
4x1 ≥ 20
x1 + 2x2≤20
A
B C
O (0, 0)

8PLANNING TECHNIQUES FOR LOGISTICS AND TRANSPORT OPERATIONS
of x1 and x2 that will be required to maximize the profit has to be determined. To do this, from
the data tab in excel, solver option has to be selected and from the solver window appearing in
the excel sheet, the target cell will be selected as the cell below the total profit in the figure
below (cell E25). In the option “By Changing Cells”, cells which will determine the numbers of
x1 and x2 required to maximize the profit (cells C17 and D17) will be selected. The constraints
will be typed in the constraints section. Then by clicking on solve, excel solver will give the
results for the required number of units to maximize the profit as well as the maximum profit.
Figure 2: Formulae for Excel Solver
From figure 3, it can be seen that 5 units of x1 and 7.5 units of x2 will be required to
maximize the profit and the maximum profit will be $95.
Figure 3: Results from Excel Solver