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If a linear programming problem has a solution, then it must occur at a _____ of the set of feasible solutions.

Short Answer

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Vertex

Step by step solution

01

Understanding Linear Programming

In the context of linear programming, a target function (either to be maximized or minimized) is determined based on given constraints. This function is evaluated over the feasible region which include the set of all possible solutions that satisfy the constraints.
02

Fundamental Theorem of Linear Programming

According to this theorem, if a solution exists to a linear programming problem, then it must exist at an extreme point (or vertex) of the feasible set. These vertices are determined by the intersection of the constraints.
03

Conclusion

Therefore the solution to a linear programming problem, if it exists, occurs at a vertex (or an extreme point) of the set of feasible solutions.

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Key Concepts

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

Feasible Region
In linear programming, the feasible region is a crucial component. It represents the set of all possible points that satisfy a problem's constraints. Think of it as a specific area on a graph where all conditions or restrictions are met. Constraints are typically inequalities that limit or define the boundaries of the feasible region.

The feasible region can be a line, a polygon, or even a multi-dimensional space, depending on the number of variables involved in the problem. Visualizing it as a shape helps you understand where feasible solutions lie.

Understanding the feasible region is essential because only within this region can we find potential solutions to maximize or minimize the target function.
Target Function
The target function, also known as the objective function, is what we aim to optimize in a linear programming problem. This function can either be a maximization (e.g., maximizing profit) or a minimization (e.g., minimizing cost) scenario.

Written mathematically, the target function is a linear expression such as \( ax + by \), where \( a \) and \( b \) are coefficients, and \( x \) and \( y \) are variables subject to the defined constraints. Evaluating this function over the feasible region allows us to determine the optimal point where our desired outcome (maximum or minimum) is achieved.

The role of the target function is pivotal as it guides the search for the best solution within the feasible region.
Extreme Point
In the context of linear programming, an extreme point, often referred to as a vertex, is a key concept. Extreme points are located at the corners or intersections of the feasible region's boundaries.

The fundamental theorem of linear programming assures us that if an optimal solution exists, it is found at one of these extreme points. This occurs because the target function's linear nature allows it to reach its maximum or minimum at the region's bounds, not inside of it.

Identifying these extreme points simplifies solving the problem, as it reduces the task to evaluating the target function only at these vertices instead of across the entire feasible region.
Constraints
Constraints in linear programming are the conditions that define the feasible region. They are usually presented as linear inequalities or equations, such as \( ax + by \leq c \).

Each constraint encompasses a boundary that limits the solution space for the variables involved. Collectively, these constraints carve out the feasible region by intersecting to form a polygonal shape, where potential solutions reside.

Understanding constraints is vital, as they directly influence the structure of the feasible region and, consequently, the location of the extreme points where feasible solutions may lie. Constraints ensure that the feasible region adheres to specific requirements necessary for problem-solving.

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Most popular questions from this chapter

Advertising A health insurance company advertises on television, on radio, and in the local newspaper. The marketing department has an advertising budget of \(\$ 42,000\) per month. A television ad costs \(\$ 1000 ,\) a radio ad costs \(\$ 200 ,\) and a newspaper ad costs \(\$ 500 .\) The department wants to run 60 ads per month and have as many television ads as radio and newspaper ads combined. How many of each type of ad can the department run each month?

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Finding Systems of Linear Equations In Exercises \(79 - 82 ,\) find two systems of linear equations that have the ordered triple as a solution. (There are many correct answers.) $$ \left( - 6 , - \frac { 1 } { 2 } , - \frac { 7 } { 4 } \right) $$

Prescriptions The numbers of prescriptions \(P\) (in thousands) filled at two pharmacies from 2009 through 2013 are shown in the table. $$ \begin{array}{|c|c|c|}\hline \text { Year } & {\text { Pharmacy A }} & {\text { Pharmacy } \mathrm{B}} \\ \hline 2009 & {19.2} & {20.4} \\ \hline 2010 & {19.6} & {20.8} \\ \hline 2011 & {20.0} & {21.1} \\ \hline 2012 & {20.6} & {21.5} \\ \hline 2013 & {21.3} & {22.0} \\ \hline\end{array} $$ (a) Use a graphing utility to create a scatter plot of the data for pharmacy A and find a linear model. Let \(t\) represent the year, with \(t=9\) corresponding to \(2009 .\) Repeat the procedure for pharmacy B. (b) Assuming that the numbers for the given five years are representative of future years, will the number of prescriptions filled at pharmacy A ever exceed the number of prescriptions filled at pharmacy B? If so, then when?

Solving a Linear Programming Problem, find the minimum and maximum values of the objective function and where they occur, subject to the indicated constraints. (For each exercise, the graph of the region determined by the constraints is provided.) $$ \begin{array}{c}{\text { Objective function: }} \\ {z=4 x+3 y} \\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {x+y \leq 5}\end{array} $$

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