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Find the maximum and the minimum values of each objective function and the values of \(x\) and \(y\) at which they occur. \(F=2 x+14 y\) subject to \(5 x+3 y \leq 34\) \(3 x+5 y \leq 30\) \(x \geq 0\) \(y \geq 0\)

Short Answer

Expert verified
Maximum value is 84 at (0, 6) and minimum value is 0 at (0, 0).

Step by step solution

01

Identify the Constraints and Objective Function

The task is to find the maximum and minimum values of the objective function: \(F = 2x + 14y\). The given constraints are: 1. \(5x + 3y \leq 34\) 2. \(3x + 5y \leq 30\) 3. \(x \geq 0\) 4. \(y \geq 0\)
02

Plot the Inequalities

Plot the inequalities on the coordinate plane by finding the intercepts and then shading the feasible region. The lines from the inequalities are: 1. \(5x + 3y = 34\) - x-intercept: \(x = \frac{34}{5}\) - y-intercept: \(y = \frac{34}{3}\) 2. \(3x + 5y = 30\) - x-intercept: \(x = 10\) - y-intercept: \(y = 6\)
03

Find the Intersection Points

Find the intersection points of the boundary lines by solving the equations simultaneously: 1. Intersection of \(5x + 3y = 34\) and \(3x + 5y = 30\). Solve these equations to get \(x\) and \(y\).
04

Evaluate the Objective Function at Boundary and Intersection Points

Evaluate the objective function \(F = 2x + 14y\) at each vertex of the feasible region.
05

Identify Maximum and Minimum Values

Determine the maximum and minimum values of the objective function from the values obtained in the previous step.
06

Solution

The critical points from the feasible region are (0,0), (0,6), (2,18/5), and (34/5,0). Evaluating the objective function at these points: - At (0,0): \(F = 2(0) + 14(0) = 0\) - At (0,6): \(F = 2(0) + 14(6) = 84\) - At (2,18/5): \(F = 2(2) + 14(18/5) = 60.8\) - At (34/5,0): \(F = 2(34/5) + 14(0) = 13.6\)Thus, the maximum value is 84 at point (0, 6), and the minimum value is 0 at point (0, 0).

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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 formula you are trying to maximize or minimize. It represents a relationship between different variables, usually in the form of a linear equation. In our exercise, the objective function is given by F = 2x + 14y The goal is to find the values of x and y that make F as large or as small as possible, considering the given constraints.
Constraints
Constraints are the conditions or limits imposed on the variables in the objective function. These constraints form a system of inequalities that define the feasible region where the solution is located. In our exercise, the constraints are: 1. 5x + 3y ≤ 34 2. 3x + 5y ≤ 30 3. x ≥ 0 4. y ≥ 0 Constraints help in drawing the feasible region on a graph, as they limit the potential values that x and y can take.
Feasible Region
The feasible region is the area on the graph where all the constraints overlap. It represents all possible solutions that satisfy the given constraints. To visualize it, you plot the lines represented by the inequalities on a coordinate system and identify the overlapping shaded area. The possible solutions are within the feasible region, and the vertex points (corners) of this region are key in finding the optimal solution.
Intersection Points
Intersection points are where the boundary lines of the constraints cross each other. These points are critical because the maximum or minimum values of the objective function often occur at these points. In our exercise, to find the intersection between 5x + 3y = 34 and 3x + 5y = 30, you solve the system of equations simultaneously. Solving these gives us a specific point (x, y) on the graph that is part of the feasible region.
Vertex Evaluation
Vertex evaluation involves calculating the value of the objective function at each vertex (corner) of the feasible region. This helps us identify which vertex yields the highest or lowest value of the objective function. For the vertices in our exercise, we evaluate 2x + 14y at points such as (0,0), (0,6), (2,18/5), and (34/5,0). The highest value at these points indicates the maximum, and the lowest value indicates the minimum.

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