/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 In Exercises \(5-8,\) find the m... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 8

In Exercises \(5-8,\) find the minimal representation of the polytope defined by the inequalities \(A \mathbf{x} \leq \mathbf{b}\) and \(\mathbf{x} \geq \mathbf{0}\) . $$ A=\left[\begin{array}{ll}{2} & {1} \\ {1} & {1} \\ {1} & {2}\end{array}\right], \mathbf{b}=\left[\begin{array}{l}{8} \\ {6} \\\ {7}\end{array}\right] $$

Short Answer

Expert verified
All constraints are essential; no redundancies found.

Step by step solution

01

Interpret the Problem

The problem asks for the minimal representation of a polytope defined by the inequalities \(A \mathbf{x} \leq \mathbf{b}\) and \(\mathbf{x} \geq \mathbf{0}\) where \(A\) and \(\mathbf{b}\) are given matrices and vector, respectively. This means we need to find all constraints that describe the feasible region without any redundancy.
02

Set Matrix Equations

Write the system of inequalities from the problem statement. The first inequality is given by the product \(A\mathbf{x} \leq \mathbf{b}\), which produces:\[2x_1 + x_2 \leq 8 \x_1 + x_2 \leq 6 \x_1 + 2x_2 \leq 7\]Along with the non-negativity constraints: \[\begin{align*}x_1 &\geq 0\x_2 &\geq 0\end{align*}\]These inequalities together define the feasible region of the polytope in the first quadrant.
03

Identify Redundant Inequalities

To simplify the inequalities, find any that can be derived from others. An inequality is redundant if its removal does not alter the feasible region of the polytope. This can be done using geometric visualization or algebraic elimination methods. Here, each inequality should be checked to see if it forms a boundary of the region.
04

Check Solutions at Intersection Points

Find the intersection points of the lines defined by the inequalities. Solve for points where two constraint lines intersect within the feasible region. For example, solve \(2x_1 + x_2 = 8\) and \(x_1 + x_2 = 6\) to find a point, then check it against all inequalities to verify if it lies within the polytope.
05

Determine Minimal Constraints

After evaluating intersections and checking redundancies, identify the minimal set of inequalities that still describe the same region. For this problem, each of the given inequalities and non-negativity constraints are essential, indicating no redundant constraints.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polytope representation
In the realm of linear programming, a polytope is a geometric concept that refers to the feasible solutions of a system of inequalities. Imagine a shape, like a polygon in 2D or a polyhedron in 3D 鈥� a polytope generalizes this to higher dimensions.
Any solution space formed by linear inequalities and equations can be characterized as a polytope. When you visualize the polytope, each vertex represents a potential solution to the inequalities.
This representation helps in understanding and solving optimization problems. It is crucial to identify the polytope as it provides a visual and mathematical means to grasp constraints and feasibility of solutions in the problem.
System of inequalities
A system of inequalities is a collection of two or more inequalities with the same set of variables. In linear programming, these inequalities define the constraints of the problem.
They illustrate boundaries beyond which the solutions cannot extend. The system is typically expressed in the standard form:
  • Matrix equation: \( A \mathbf{x} \leq \mathbf{b} \)
  • Non-negativity constraints: \( \, \mathbf{x} \geq \mathbf{0} \)
This defines a region where all solutions fit, known as the feasible region.
Understanding the system of inequalities is crucial because it forms the basis of finding eligible solutions which satisfy all imposed restrictions.
Feasible region
The feasible region is fundamentally where all possible solutions that fulfill every inequality in a system can be found. This area is bounded by the intersection of the inequalities involved.
In simpler terms, think of it as the safe zone where all conditions are satisfied. When visualized, it usually takes the shape of a polygon or polyhedron, depending on the number of variables.
The feasible region is the primary focus when solving optimization problems since it contains all the potential solutions, including the optimum one. By analyzing the feasible region, one can evaluate intersection points and boundaries to determine the optimal solution.
Redundant constraints
In any system of inequalities, constraints that do not affect the feasible region are termed redundant. These constraints do not confine the region any further than the essential constraints already do.
Identifying redundant constraints is critical because simplifying the system can lead to more efficient computations and a clearer understanding of the problem.
In practice, a constraint is redundant if you remove it and the feasible region remains unchanged. Geometric interpretation or algebraic techniques can help recognize such constraints. By evaluating each inequality, you filter out those that do not define the polytope's boundary. This streamlines both analysis and solution processes of optimization problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 9 and \(10,\) mark each statement True or False. Justify each answer. a. If \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\) are in \(\mathbb{R}^{n}\) and if the set \(\left\\{\mathbf{v}_{1}-\mathbf{v}_{2}, \mathbf{v}_{3}-\mathbf{v}_{2}, \ldots,\right.\) \(\mathbf{v}_{p}-\mathbf{v}_{2} \\}\) is linearly dependent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) affinely dependent. (Read this carefully.) b. If \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\) are in \(\mathbb{R}^{n}\) and if the set of homogeneous forms \(\left\\{\tilde{\mathbf{v}}_{1}, \ldots, \tilde{\mathbf{v}}_{p}\right\\}\) in \(\mathbb{R}^{n+1}\) is linearly independent, then \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is affinely dependent. c. A finite set of points \(\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{k}\right\\}\) is affinely dependent if there exist real numbers \(c_{1}, \ldots, c_{k},\) not all zero, such that \(c_{1}+\cdots+c_{k}=1\) and \(c_{1} \mathbf{v}_{1}+\cdots+c_{k} \mathbf{v}_{k}=\mathbf{0} .\) d. If \(S=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\right\\}\) is an affinely independent set in \(\mathbb{R}^{n}\) and if \(\mathbf{p}\) in \(\mathbb{R}^{n}\) has a negative barycentric coordinate determined by \(S,\) then \(\mathbf{p}\) is not in aff \(S .\) e. If \(\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}, \mathbf{a},\) and \(\mathbf{b}\) are in \(\mathbb{R}^{3}\) and if a ray \(\mathbf{a}+t \mathbf{b}\) for \(t \geq 0\) intersects the triangle with vertices \(\mathbf{v}_{1}, \mathbf{v}_{2},\) and \(\mathbf{v}_{3}\) then the barycentric coordinates of the intersection point are all nonnegative.

In Exercises \(1-4,\) write \(\mathbf{y}\) as an affine combination of the other points listed, if possible. $$ \mathbf{v}_{1}=\left[\begin{array}{l}{1} \\ {2} \\ {0}\end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{r}{2} \\ {-6} \\ {7}\end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{l}{4} \\ {3} \\ {1}\end{array}\right], \mathbf{y}=\left[\begin{array}{r}{-3} \\ {4} \\ {-4}\end{array}\right] $$

In Exercises 7 and \(8,\) find the barycentric coordinates of \(\mathbf{p}\) with respect to the affinely independent set of points that precedes it. $$ \left[\begin{array}{r}{0} \\ {1} \\ {-2} \\\ {1}\end{array}\right],\left[\begin{array}{l}{1} \\ {1} \\ {0} \\\ {2}\end{array}\right],\left[\begin{array}{r}{1} \\ {4} \\ {-6} \\\ {5}\end{array}\right], \mathbf{p}=\left[\begin{array}{r}{-1} \\ {1} \\ {-4} \\\ {0}\end{array}\right] $$

In Exercises 11 and \(12,\) mark each statement True or False. Justify each answer. a. The cubic B茅zier curve is based on four control points. b. Given a quadratic B茅zier curve \(\mathbf{x}(t)\) with control points \(\mathbf{p}_{0}, \mathbf{p}_{1},\) and \(\mathbf{p}_{2},\) the directed line segment \(\mathbf{p}_{1}-\mathbf{p}_{0}\) (from \(\mathbf{p}_{0}\) to \(\mathbf{p}_{1} )\) is the tangent vector to the curve at \(\mathbf{p}_{0}\) . c. When two quadratic B茅zier curves with control points \(\left\\{\mathbf{p}_{0}, \mathbf{p}_{1}, \mathbf{p}_{2}\right\\}\) and \(\left\\{\mathbf{p}_{2}, \mathbf{p}_{3}, \mathbf{p}_{4}\right\\}\) are joined at \(\mathbf{p}_{2},\) the combined B茅zier curve will have \(C^{1}\) continuity at \(\mathbf{p}_{2}\) if \(\mathbf{p}_{2}\) is the midpoint of the line segment between \(\mathbf{p}_{1}\) and \(\mathbf{p}_{3} .\)

In Exercises \(15-20,\) write a formula for a linear functional \(f\) and specify a number \(d,\) so that \([f : d]\) is the hyperplane \(H\) described in the exercise. Let \(H\) be the column space of the matrix \(B=\left[\begin{array}{rr}{1} & {0} \\\ {5} & {2} \\ {-4} & {-4}\end{array}\right]\) That is, \(H=\operatorname{Col} B\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.