Introduction To Management Science, Global Edition
Introduction To Management Science, Global Edition
13th Edition
ISBN: 9781292263045
Author: Bernard W. Taylor III
Publisher: PEARSON
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Chapter 2, Problem 1P

a.

Summary Introduction

To formulate: A linear programming model.

Introduction: The linear programming model is used to minimize the cost and maximize the profit through algebraic and graphical presentation using different variables and satisfying them.

a.

Expert Solution
Check Mark

Explanation of Solution

Given information:

Company TM wants to maximize cakes and loaf of breads and is willing to sell a cake for $10 and a loaf of bread for $6.

Maximize: Z=10x+6y

Subject to:

3x+8y2045x+30y180x,y0

b.

Summary Introduction

To determine: Show the graphical analysis of the model.

Introduction: The linear programming model is used to minimize the cost and maximize the profit through algebraic and graphical presentation using different variables and satisfying them.

b.

Expert Solution
Check Mark

Explanation of Solution

Given information:

Company TM wants to maximize cakes and loaf of breads and is willing to sell a cake for $10 and a loaf of bread for $6.

The coordinates for the above equations are:

Introduction To Management Science, Global Edition, Chapter 2, Problem 1P , additional homework tip  1

Table (1)

By solving the above equations, the values of x,y are as follows:

Introduction To Management Science, Global Edition, Chapter 2, Problem 1P , additional homework tip  2

Table (2)

The graphical analysis is as follows:

Introduction To Management Science, Global Edition, Chapter 2, Problem 1P , additional homework tip  3

Fig (1)

A feasible solution (ABCD) where objective function is at its maximum point is known as optimal solution. At this point, no other feasible and better solution exists. This point exists either at the top or at bottom as per the requirement of objective function and its constraints.

Therefore, the optimal solution is point C (4,0)

Calculation of Z at (4,0)

Z=10(4)+6(0)=40+0=40

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