Discuss about the Application of Linear Algebra in Solving Real Life Problem.

Linear algebra has a wide variety of applications in solving real-life problems such as Image compression, Markov chains, Estimating page relevance for search engines, Computer graphics, Machine learning, Least squares fitting in statistics, and linear programming. This paper examines how linear algebra can be applied to solve a real problem in the production setup of a factory. The particular problem solved in this paper involves the determination of linear programming constraints and variables by the use of linear algebra. In order to solve the linear programming problems, linear algebra is used to formulate the various variables and constraints with the view of simplifying the word problem into mathematical representation that is easy to solve.

Linear programming can be employed by the personnel management to work out the problems related to employee recruitment, training, selection, and deployment. The concept helps in determining the minimum number of workers that can work in a shift to ensure the production schedule is met within the stipulated time (Anieting, 2013). In addition, linear programming can assist in the management of an inventory in a factory in relation to the finished products and the raw materials. Linear programming thus, helps in determining the minimum cost of inventory in relation to the constraints of space and cost. In the current problem, linear programming is used in solving a blending problem in the process of production. Blending involves the mixing of different raw materials in a ratio that minimises the cost of production and maximises the revenue or profit. In this particular problem, linear algebra is employed in the formulation of the various variables and constraints to help in simplifying the problem by converting it into a mathematical representation that is easy to solve by assigning letters to the unknown variables. The variables are then converted into linear inequalities that can easily be solved using graphical method.

This is a real life problem that majorly affect producers in working out the best mix in coming up with a blend of a product. In this problem, a producer of candies intends to produce chocolate coated with either cherries or mints. The decision to come up with the best blend that minimises the production cost and maximises the revenue depend on the cost of the individual components and the most probable mixing ratio that takes care of the cost of the raw materials. In order to solve the problem, variables must be established from the constraints provided by the producer.

A toffee producing company has 130 kilograms of chocolate coated with cherries and 170 kilograms of chocolate coated with mints in to distribute. The company sells these products in the form of two separate mixtures. The first mixture is intended to contain half mints and half cherries by mass and is intended to be sold at $2 per kilogram. The second mixture is intended to contain two thirds of mints and a third of cherries by mass and is intended to be sold at $1.25 per kilogram. Determine the mass of each mixture that the company should sell maximise their profit.

In order to simplify the problem, letters will be used to represent the variables. The variables will then be used to formulate linear algebraic equations. A dependent variable will then be established to formulate a linear programming equation that will optimise the variables to maximise sales. The constraints and variables will be related to one another using linear inequalities. Once the problem has been reduced into a mathematical representation, graphical method will be used to optimise the value of the objective equation. In the graphical method, coordinates arising from the linear inequalities derived will be plotted on a Cartesian plane and the maximum value of the objective function will determine the solution to the linear inequalities formulated.

This problem requires the knowledge of linear algebra to come up with its solution. Linear algebra combines linear functions and vectors to solve complicated problems (Wilkinson, Reinsch, & Bauer, 2013). To simplify the problem,

Let A represent the blend of half cherries and half mints.

Let B represent the blend of a third of cherries and two thirds of mints.

Let x be the number of kilograms of A to be manufactured

Let y be the number of kilograms of B to be manufactured

Let the he profit function be represented as

Z = 2x + 1.25 y

Since each kilogram of A contains half kilograms of cherries and each kilogram of B contains a third kilogram of cherries, the total number of kilograms of cherries used in both blend is represented as

1/2 x + 1/3 y

In a similar manner, the total number of kilograms of mints used in both blend is:

1/2 x + 2/3 y

Therefore, since the company can utilise at most 130 kilograms of cherries and 170 kilograms of mints, the other constraints can be formulated as:

1/2 x + 1/3 y ≤ 130

1/2 x + 2/3 y ≤ 170

In addition, we there should be other basic constraints as follows;

x ≥ 0

y ≥ 0

For that reason, the problem can be formulated as indicated below so solve for x and y that maximise the profit using the subject Z = 2x + 1.25 y

1/2 x + 1/3 y ≤ 130

1/2 x + 2/3 y ≤ 170

x ≥ 0

y ≥ 0

In order to solve the problem, the technique of linear programming is used.

Step 1:

We begin by indicating the feasible region of the problem and locating the resultant extreme points:

Vertex |
Lines Through Vertex |
Value of Objective |

(180,120) |
1/2x +1/3y = 130; 1/2x + 2/3y = 170 |
510 |

(260,0) |
1/2x + 1/3y = 130; y = 0 |
520 Maximum |

(0,255) |
1/2x + 2/3y = 170; x = 0 |
318.75 |

(0,0) |
x = 0; y = 0 |
0 |

The four points in the table will be plotted on a Cartesian plane to find out where the value of z is a maximum. The plotting is done in the next section followed by an explanation of the solution.

The coordinates above can be plotted on a Cartesian plane to come up with the probable maximum value of Z. The graph below shows the solution of the linear programming problem.

Figure 1 A graphical solution of the Linear Programming problem

Once the graph is plotted and the values obtained, the values of the objective function that gives the maximum value of the variables are established. From the graph, the highest value of z is 520 and this corresponds to the point (260, 0). This point gives the optimal solution to the problem. Therefore, the manufacturer of the toffee is likely to maximise his profits of $520 when he manufactures 260 kilograms of option A and 0 kilograms of option B. Since each kilogram of option A contains half kilograms of cherries and each kilogram of B contains a third kilogram of cherries, the manufacturer maximises profit by selling half kilograms of cherries. The implication of this solution is that for the manufacturer to maximise his revenue, he should not sell or produces option B of the candies since it will not fetch any profit. This means the company may is likely to run at a loss should it venture into this option.

In conclusion, linear algebra has a number of applications in real life including Image compression, Markov chains, Estimating page relevance for search engines, Computer graphics, Machine learning, Least squares fitting in statistics, and linear programming. The particular problem solved in the paper demonstrates how linear algebra can be used to solve problems related to linear programming. In the industry, linear programming thus, helps in determining the minimum cost of inventory in relation to the constraints of space and cost. It can also be employed in determining the minimum number of workers that can work in a shift to ensure the production schedule is met within the stipulated time. In addition, to solve the problems related to employee recruitment, training, selection, and deployment, the concept of linear programming becomes important for the personnel management. In order to solve all these problems, linear algebra is used to formulate the various variables and constraints with the view of simplifying the word problem into mathematical representation that is easy to solve.

Nahmias, S., & Cheng, Y. (2009). Production and operations analysis (Vol. 6). New York: McGraw-Hill.

A.E. Anieting, A. (2013). Application of Linear Programming Technique in the Determination of Optimum Production Capacity. IOSR Journal Of Mathematics, 5(6), 62-65. https://dx.doi.org/10.9790/5728-0566265

Wilkinson, J., Reinsch, C., & Bauer, F. (2013). Linear algebra (2nd ed.). Berlin: Springer-Verlag.

Discuss about the Application of Linear Algebra in Solving Real Life Problem.

Linear algebra has a wide variety of applications in solving real-life problems such as Image compression, Markov chains, Estimating page relevance for search engines, Computer graphics, Machine learning, Least squares fitting in statistics, and linear programming. This paper examines how linear algebra can be applied to solve a real problem in the production setup of a factory. The particular problem solved in this paper involves the determination of linear programming constraints and variables by the use of linear algebra. In order to solve the linear programming problems, linear algebra is used to formulate the various variables and constraints with the view of simplifying the word problem into mathematical representation that is easy to solve.

Linear programming can be employed by the personnel management to work out the problems related to employee recruitment, training, selection, and deployment. The concept helps in determining the minimum number of workers that can work in a shift to ensure the production schedule is met within the stipulated time (Anieting, 2013). In addition, linear programming can assist in the management of an inventory in a factory in relation to the finished products and the raw materials. Linear programming thus, helps in determining the minimum cost of inventory in relation to the constraints of space and cost. In the current problem, linear programming is used in solving a blending problem in the process of production. Blending involves the mixing of different raw materials in a ratio that minimises the cost of production and maximises the revenue or profit. In this particular problem, linear algebra is employed in the formulation of the various variables and constraints to help in simplifying the problem by converting it into a mathematical representation that is easy to solve by assigning letters to the unknown variables. The variables are then converted into linear inequalities that can easily be solved using graphical method.

This is a real life problem that majorly affect producers in working out the best mix in coming up with a blend of a product. In this problem, a producer of candies intends to produce chocolate coated with either cherries or mints. The decision to come up with the best blend that minimises the production cost and maximises the revenue depend on the cost of the individual components and the most probable mixing ratio that takes care of the cost of the raw materials. In order to solve the problem, variables must be established from the constraints provided by the producer.

A toffee producing company has 130 kilograms of chocolate coated with cherries and 170 kilograms of chocolate coated with mints in to distribute. The company sells these products in the form of two separate mixtures. The first mixture is intended to contain half mints and half cherries by mass and is intended to be sold at $2 per kilogram. The second mixture is intended to contain two thirds of mints and a third of cherries by mass and is intended to be sold at $1.25 per kilogram. Determine the mass of each mixture that the company should sell maximise their profit.

In order to simplify the problem, letters will be used to represent the variables. The variables will then be used to formulate linear algebraic equations. A dependent variable will then be established to formulate a linear programming equation that will optimise the variables to maximise sales. The constraints and variables will be related to one another using linear inequalities. Once the problem has been reduced into a mathematical representation, graphical method will be used to optimise the value of the objective equation. In the graphical method, coordinates arising from the linear inequalities derived will be plotted on a Cartesian plane and the maximum value of the objective function will determine the solution to the linear inequalities formulated.

This problem requires the knowledge of linear algebra to come up with its solution. Linear algebra combines linear functions and vectors to solve complicated problems (Wilkinson, Reinsch, & Bauer, 2013). To simplify the problem,

Let A represent the blend of half cherries and half mints.

Let B represent the blend of a third of cherries and two thirds of mints.

Let x be the number of kilograms of A to be manufactured

Let y be the number of kilograms of B to be manufactured

Let the he profit function be represented as

Z = 2x + 1.25 y

Since each kilogram of A contains half kilograms of cherries and each kilogram of B contains a third kilogram of cherries, the total number of kilograms of cherries used in both blend is represented as

1/2 x + 1/3 y

In a similar manner, the total number of kilograms of mints used in both blend is:

1/2 x + 2/3 y

Therefore, since the company can utilise at most 130 kilograms of cherries and 170 kilograms of mints, the other constraints can be formulated as:

1/2 x + 1/3 y ≤ 130

1/2 x + 2/3 y ≤ 170

In addition, we there should be other basic constraints as follows;

x ≥ 0

y ≥ 0

For that reason, the problem can be formulated as indicated below so solve for x and y that maximise the profit using the subject Z = 2x + 1.25 y

1/2 x + 1/3 y ≤ 130

1/2 x + 2/3 y ≤ 170

x ≥ 0

y ≥ 0

In order to solve the problem, the technique of linear programming is used.

Step 1:

We begin by indicating the feasible region of the problem and locating the resultant extreme points:

Vertex |
Lines Through Vertex |
Value of Objective |

(180,120) |
1/2x +1/3y = 130; 1/2x + 2/3y = 170 |
510 |

(260,0) |
1/2x + 1/3y = 130; y = 0 |
520 Maximum |

(0,255) |
1/2x + 2/3y = 170; x = 0 |
318.75 |

(0,0) |
x = 0; y = 0 |
0 |

The four points in the table will be plotted on a Cartesian plane to find out where the value of z is a maximum. The plotting is done in the next section followed by an explanation of the solution.

The coordinates above can be plotted on a Cartesian plane to come up with the probable maximum value of Z. The graph below shows the solution of the linear programming problem.

Figure 1 A graphical solution of the Linear Programming problem

Once the graph is plotted and the values obtained, the values of the objective function that gives the maximum value of the variables are established. From the graph, the highest value of z is 520 and this corresponds to the point (260, 0). This point gives the optimal solution to the problem. Therefore, the manufacturer of the toffee is likely to maximise his profits of $520 when he manufactures 260 kilograms of option A and 0 kilograms of option B. Since each kilogram of option A contains half kilograms of cherries and each kilogram of B contains a third kilogram of cherries, the manufacturer maximises profit by selling half kilograms of cherries. The implication of this solution is that for the manufacturer to maximise his revenue, he should not sell or produces option B of the candies since it will not fetch any profit. This means the company may is likely to run at a loss should it venture into this option.

In conclusion, linear algebra has a number of applications in real life including Image compression, Markov chains, Estimating page relevance for search engines, Computer graphics, Machine learning, Least squares fitting in statistics, and linear programming. The particular problem solved in the paper demonstrates how linear algebra can be used to solve problems related to linear programming. In the industry, linear programming thus, helps in determining the minimum cost of inventory in relation to the constraints of space and cost. It can also be employed in determining the minimum number of workers that can work in a shift to ensure the production schedule is met within the stipulated time. In addition, to solve the problems related to employee recruitment, training, selection, and deployment, the concept of linear programming becomes important for the personnel management. In order to solve all these problems, linear algebra is used to formulate the various variables and constraints with the view of simplifying the word problem into mathematical representation that is easy to solve.

Nahmias, S., & Cheng, Y. (2009). Production and operations analysis (Vol. 6). New York: McGraw-Hill.

A.E. Anieting, A. (2013). Application of Linear Programming Technique in the Determination of Optimum Production Capacity. IOSR Journal Of Mathematics, 5(6), 62-65. https://dx.doi.org/10.9790/5728-0566265

Wilkinson, J., Reinsch, C., & Bauer, F. (2013). Linear algebra (2nd ed.). Berlin: Springer-Verlag.

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