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Chapter 7: Problem 12

Optimize \(5 x+3 y\) subject to $$ \begin{array}{r} 1.2 x+0.6 y \leq 24 \\ 2 x+1.5 y \leq 80 \end{array} $$

Short Answer

Expert verified
The maximum value of the objective function is 200 at \((40, 0)\).

Step by step solution

01

Understand the Problem

We need to optimize (maximize or minimize) the objective function, which is \(5x + 3y\), subject to given linear inequalities: \(1.2x + 0.6y \leq 24\) and \(2x + 1.5y \leq 80\). These constraints will form a feasible region on a coordinate plane.
02

Identify and Graph the Constraints

Graph the constraints on a coordinate plane. Transform the inequalities into equalities to find the boundary lines:1. From \(1.2x + 0.6y = 24\), solving for \(y\) gives \(y = 40 - 2x\).2. From \(2x + 1.5y = 80\), solving for \(y\) gives \(y = \frac{80 - 2x}{1.5}\).Then, shade the region that satisfies both inequalities (below both lines).
03

Find the Intersection Points

Calculate the intersection points of the boundary lines and where they intersect the axes:- Set \(1.2x + 0.6y = 24\) equal to \(2x + 1.5y = 80\) to find their intersection. Solving the system gives \(x = 15\) and \(y = 10\). - Find the intercepts by solving each equation when \(x = 0\) and when \(y = 0\). 1. At \(x = 0\), \(0.6y = 24\) gives \(y = 40\). 2. At \(y = 0\), \(1.2x = 24\) gives \(x = 20\). 3. For \(2x + 1.5y = 80\), when \(x = 0\), \(1.5y = 80\) gives \(y = \frac{160}{3}\), and when \(y = 0\), \(2x = 80\) gives \(x = 40\).
04

Define and Use the Feasible Region

Identify the feasible region from the graph, which is the area shared by both inequalities. It should be a bounded region or a shape such as a quadrilateral formed by the intersections and intercept points (solutions of the lines and axes boundaries).
05

Evaluate the Objective Function at Vertex Points

The optimal solution for linear programming occurs at the vertices of the feasible region.1. Calculate the objective function \(5x + 3y\) at each vertex point obtained: - At \((15, 10)\): \(5(15) + 3(10) = 105\). - At \((20, 0)\): \(5(20) + 3(0) = 100\). - At \((0, 40)\): \(5(0) + 3(40) = 120\). - At \((40, 0)\), \(5(40) + 3(0) = 200\).2. Determine the maximum value.
06

Determine the Optimal Solution

After evaluating the objective function at all vertices, the point \((40, 0)\) provides the maximum value for the function, which is 200. This is the optimal solution for the given linear programming problem.

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Key Concepts

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

Objective Function
In linear programming, the objective function is the mathematical expression that we aim to optimize, meaning we either want to maximize or minimize it.
For example, in the given problem, the objective function is given by \(5x + 3y\).
This function is what we're interested in optimizing subject to a set of constraints.
  • "Maximize" means we want to find the largest possible value of the function.
  • "Minimize" means we seek the smallest possible value.
The objective function often represents some quantity like profit or cost that needs optimization.
By analyzing this function at specific points of a feasible region (which we'll learn about shortly), we can determine the optimal solution for a particular linear programming problem.
Feasible Region
The feasible region is the set of all possible points that satisfy the constraints of a linear programming problem. These constraints are typically given in the form of inequalities and, when plotted on a coordinate plane, will form a region.
For instance, in our example, the constraints \(1.2x + 0.6y \leq 24\) and \(2x + 1.5y \leq 80\) create a particular area when graphed.
This region is often a polygon, like a triangle or quadrilateral, made up of the intersection of inequality boundaries.
  • Each point in this region is a potential solution that meets all given constraints.
  • The feasible region is crucial because it visually pinpoints where the optimal solution lies.
  • An optimal solution will always be found at one of the vertices of the feasible region.
Understand that not all regions will be feasible. If no intersection exists within the constraints, then a feasible region is not present, indicating no viable solution.
Intersection Points
Intersection points are where the boundary lines of constraints intersect or meet each other.
These points can be crucial in finding the vertices of the feasible region, where the potential optimal solutions live.
In our problem, setting the equations \(1.2x + 0.6y = 24\) and \(2x + 1.5y = 80\) equal allows us to solve for their intersection, which can be found using basic algebra.
  • Solving these simultaneously gives an intersection at \((15, 10)\).
  • These points are calculated similarly by solving for when \(x = 0\) or \(y = 0\), also known as intercepts.
  • For example, at \(x = 0\), you solve for \(y\); and at \(y = 0\), solve for \(x\), which further helps determine other points along the lines.
Identify all intersection points and key intercepts to fully understand the shape of the feasible region, leading to the location of the optimal solution.
Constraint Graphing
Constraint graphing involves plotting the constraints expressed as inequalities to visualize the feasible region.
This can be done by first converting inequalities into equalities to graph their boundary lines.Start by identifying the constraints, then draw these lines on a coordinate plane.
For example, transform \(1.2x + 0.6y \leq 24\) into \(1.2x + 0.6y = 24\) and \(2x + 1.5y \leq 80\) into \(2x + 1.5y = 80\).
  • Next, calculate intercepts by setting \(x = 0\) and \(y = 0\) to find where lines hit the axes.
  • Shade the area on the graph that satisfies all constraints; this shaded area represents the feasible region.
The use of graphing effectively illustrates the problem's requirements, making the process of identifying the optimal solution more manageable.

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