Introduction To Operations Research Hillier 8th Edition Solution 18 ^HOT^

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How to Solve Problem 18 in Chapter 18 of Introduction to Operations Research by Hillier and Lieberman

Introduction to Operations Research is a classic textbook that covers various topics in operations research, such as linear programming, network models, simulation, and optimization theory. The book is written by Frederick Hillier and Gerald Lieberman, who are both renowned experts in the field. The book has been updated and revised several times to reflect the latest developments and applications of operations research.

In this article, we will focus on one of the problems in Chapter 18 of the book, which deals with classical optimization theory. This chapter introduces the concepts and methods of unconstrained optimization, constrained optimization, and Lagrange multipliers. The problem we will solve is Problem 18 in Problem Set 18.2A, which asks us to find the minimum value of the function f(x,y) = x^4 + y^4 - 4xy + 1, subject to the constraint x + y = 2.

To solve this problem, we will use the method of Lagrange multipliers, which is a technique for finding the extrema of a function subject to one or more constraints. The basic idea is to construct a new function called the Lagrangian, which combines the original function and the constraints using some auxiliary variables called Lagrange multipliers. Then, we find the stationary points of the Lagrangian by setting its partial derivatives equal to zero. Finally, we check which of these points satisfy the original constraints and evaluate the original function at these points to find the extrema.

Let's apply this method to our problem. First, we construct the Lagrangian as follows:

L(x,y,Î) = f(x,y) - Î(g(x,y) - c)

where f(x,y) is the original function, g(x,y) is the constraint function, c is a constant, and Î is the Lagrange multiplier. In our case, we have:

L(x,y,Î) = x^4 + y^4 - 4xy + 1 - Î(x + y - 2)

Next, we find the stationary points of L by setting its partial derivatives equal to zero:

âL/âx = 4x^3 - 4y - Î = 0

âL/ây = 4y^3 - 4x - Î = 0

âL/âÎ = x + y - 2 = 0

Solving this system of equations is not trivial, but we can use some algebraic tricks to simplify it. First, we add the first two equations and get:

4x^3 + 4y^3 - 8x - 8y = 0

Then, we divide both sides by 4 and factor out x + y:

(x + y)(x^2 - xy + y^2 - 2) = 0

Since x + y = 2 from the third equation, we can substitute it into the second factor and get:

(x^2 - xy + y^2 - 2)(2) = 0

x^2 - xy + y^2 - 2 = 0

This is a quadratic equation in terms of xy. We can solve it using the quadratic formula:

xy = (-(-1) Â â((-1)^2 - 4(1)(-2)))/(2(1))

xy = (1 Â â9)/2

xy = (1 Â 3)/2

xy = 2 or xy = -1

Now we have two possible values for xy. For each value, we can plug it into x + y = 2 and get a pair of values for x and y. For xy = 2, we have:

x + y = 2

x(2/x) + y = 2

2 + y = aa16f39245