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Inequalities in Linear Programming Model

by Nkomazana Mthokoe
(Harare, Zimbabwe)










































A diet contains at least 16 units of carbohydrates and 20 units of protein. Food A contains 2 units of carbohydrates and 4 units of protein. Food B contains 2 units of carbohydrates and 1 unit of protein. If Food A costs $1.20 per unit and Food B costs $0.80, how many units of each should be purchased in order to minimize cost?

Comments for Inequalities in Linear Programming Model

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Sep 13, 2012
Minimize Cost
by: Staff


Answer:


Part I


The problem statement uses the word “units” to stand for two entirely different things: 1) the quantity of carbohydrates and protein, and 2) the quantity of food A and food B.


To clearly identify the difference between 1) and 2), I’m replacing the word “units” for 1) with the word “parts”.

This change will not affect the answer to your question.

However, it should make the mathematics a little easier to follow.


Original problem statement

A diet contains at least 16 UNITS of carbohydrates and 20 UNITS of protein. Food A contains 2 UNITS of carbohydrates and 4 UNITS of protein. Food B contains 2 UNITS of carbohydrates and 1 UNIT of protein. If Food A costs $1.20 per unit and Food B costs $0.80, how many units of each should be purchased in order to minimize cost?


Revised problem statement

A diet contains at least 16 PARTS of carbohydrates and 20 PARTS of protein. Food A contains 2 PARTS of carbohydrates and 4 PARTS of protein. Food B contains 2 PARTS of carbohydrates and 1 PART of protein. If Food A costs $1.20 per unit and Food B costs $0.80, how many units of each should be purchased in order to minimize cost?



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Definitions:

         x = units (quantity) of food A purchased

         y = units (quantity) of food B purchased


Equation for Cost of Food :

         Cost = total cost of food

         Cost = cost of food A + cost of food B

         cost of food A = $1.20 per unit

         cost of food B = $0.80 per unit

         Cost = $1.20x + $0.80y


Requirement for carbohydrates:

         carbohydrates ≥ 16 parts

         (remember, I changed the word used to represent the quantity of carbohydrates from “units” to “parts” to make the problem easier to follow)

         One unit (quantity) of food A purchased = 2 parts of carbohydrates

         One unit (quantity) of food B purchased = 2 parts of carbohydrates

         (carbohydrates from food A) + (carbohydrates from food B) ≥ 16 parts

         2x + 2y ≥ 16

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Sep 13, 2012
Minimize Cost
by: Staff


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Part II


Requirement for protein :


         protein ≥ 20 parts

         (remember, I changed the word used to represent the quantity of protein from “units” to “parts” to make the problem easier to follow)

         One unit (quantity) of food A purchased = 4 parts of protein

         One unit (quantity) of food B purchased = 1 part of protein

         (carbohydrates from food A) + (carbohydrates from food B) ≥ 16 parts

         4x + 1y ≥ 20



Mathematically, the Problem is:

         Minimize the objective cost function:

             Cost = $1.20x + $0.80y

         Subject to the following constraints:

             2x + 2y ≥ 16

             4x + 1y ≥ 20

             x ≥ 0

             y ≥ 0


To solve this problem:

         1. Plot the restrictions on x-y coordinates

         2. Compute the Cost Function at the corner points of the unbounded feasible region


1. Plot the restrictions :

         2x + 2y ≥ 16

             (the red line and all the area above the red line)

         4x + 1y ≥ 20

             (the green line and all the area above the green line)

         x ≥ 0

             (the blue line and all the area to the right of the vertical blue line)

---------------------------------------

Sep 13, 2012
Minimize Cost
by: Staff


---------------------------------------

Part III

         y ≥ 0

             (the pink line and all the area above the horizontal pink line)


         Un-Bounded Feasible Region

             (shaded yellow area)





 Math – graph of unbounded feasible region





2. Compute the Cost Function at the corner points of the unbounded feasible region :






 Math – graph of feasible region & corner points






         The corner points are :

             (0,20), (4,4), (8,0)


         Find the minimum value of the cost function using the x-y values at the corner points.

             Cost = $1.20x + $0.80y


             Corner point (0,20)

             Cost = $1.20 * 0 + $0.80 * 20

             Cost = 0 + $16.00

             Cost = $16.00



             Corner point (4,4)

             Cost = $1.20 * 4 + $0.80 * 4

             Cost = $4.80 + $3.20

             Cost = $8.00, minimum cost


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Sep 13, 2012
Minimize Cost
by: Staff


---------------------------------------

Part IV




             Corner point (8,0)

             Cost = $1.20 * 8 + $0.80 * 0

             Cost = $9.60 + 0

             Cost = $9.60


The Final Answer

          Minimum Cost is: $8

          When,

          Food A = 4 units purchased

          Food B = 4 units purchased



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Check the results:

          Do the number of units purchased of Food A and Food B meet the restrictions imposed by the following two inequalities?

             2x + 2y ≥ 16

             4x + 1y ≥ 20


          x (Food A) = 4 units

          y (Food B) = 4 units


          2x + 2y ≥ 16

          2*4 + 2*4 ≥ 16

          8 + 8 ≥ 16

          16 ≥ 16, True



          4x + 1y ≥ 20

          4*4 + 1*4 ≥ 20

          16 + 4 ≥ 20

          20 ≥ 20, True




Thanks for writing.

Staff
www.solving-math-problems.com



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