Problem 6 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 5-6: Sketch the region \(R\) determined by the given constraints, and label its vertices. Find the minimum value of \(C\) on \(R\). $$ \begin{array}{ll} C=6 x+y ; & y \geq 0 \\ 3 x+y \geq 3, & x+5 y \leq 15, \quad 2 x+y \leq 12 \end{array} $$

Step-by-Step Solution

Answer

The minimum value of \( C \) is 3 at the vertex (0, 3).

Step 1: Understand the Constraints

The given inequalities define the region of interest. These are: 1. \( y \geq 0 \), ensuring all points are above the x-axis.2. \( 3x + y \geq 3 \), which can be rewritten as \( y \geq -3x + 3 \).3. \( x + 5y \leq 15 \), which can be rewritten as \( y \leq \frac{15-x}{5} \).4. \( 2x + y \leq 12 \), which can be rewritten as \( y \leq -2x + 12 \).

Step 2: Graph the Inequalities

Graph each inequality on the coordinate plane:- Graph the line for \( y = -3x + 3 \) and shade above it.- Graph the line for \( y = \frac{15-x}{5} \) and shade below it.- Graph the line for \( y = -2x + 12 \) and shade below it.Also plot \( y \geq 0 \), which is above the x-axis.The feasible region \( R \) is where all shaded areas overlap.

Step 3: Identify the Vertices of Region R

Find intersection points of the lines:1. Intersection of \( y = -3x + 3 \) and \( y = \frac{15-x}{5} \): - Solve the equations: \(-3x + 3 = \frac{15-x}{5}\) - Solving gives intersection at \( (0, 3) \).2. Intersection of \( y = \frac{15-x}{5} \) and \( y = -2x + 12 \): - Solve the equations: \( \frac{15-x}{5} = -2x + 12 \) - Solving gives intersection at \( (5, 2) \).3. Intersection of \( y = -3x + 3 \) and \( y = -2x + 12 \): - Solve the equations: \(-3x + 3 = -2x + 12 \) - Solving gives intersection at \( (9, -15) \), but since \( y \geq 0 \), not a vertex of \( R \).

Step 4: Evaluate C at Each Vertex

For each vertex of the feasible region, calculate the value of \( C = 6x + y \).1. At \( (0, 3) \): \( C = 6(0) + 3 = 3 \).2. At \( (5,2) \): \( C = 6(5) + 2 = 32 \).3. The rest do not contribute additional valid vertices.

Step 5: Determine the Minimum Value of C

Among the calculated values of \( C \), the smallest value is 3 at the vertex \( (0, 3) \).

Step 6: Conclusion

The region \( R \) and its vertices must be sketched to visualize the overlaps. The minimum value of \( C = 6x + y \) on region \( R \) is attained at the vertex \( (0, 3) \), and this minimum value is 3.

Key Concepts

Feasible Region

In linear programming, the feasible region is a key concept. It represents all the possible solutions that satisfy a system of inequalities. To find this region, you graph each inequality on a coordinate plane. This involves drawing their boundary lines and shading the appropriate side of each line.

The feasible region is where all of these shaded areas overlap. It is essentially the set of all points that meet every given constraint in the problem. Points inside or on the boundary of the feasible region are possible solutions to the problem.

In our original problem, the feasible region is determined by:

• The line where all points lie above or on it: \( y \geq 0 \)
• The line with the constraint: \( y \geq -3x + 3 \)
• Another line where the shade lies below it: \( y \leq \frac{15-x}{5} \)
• Finally, shaded below this line: \( y \leq -2x + 12 \)

Understanding these shaded overlaps helps us visualize all possible (x, y) pairs that satisfy the constraints.

Vertices of a Polygon

Vertices play a crucial role in linear programming as they are points where boundary lines intersect in the context of inequalities. These points mark the corners of the feasible region, which, in many cases, forms a polygon.

To find the vertices of this polygon, you must solve pairs of equations where two lines intersect. This involves some algebra to ensure each vertex obeys all constraints.

In our example, vertices are located at:

• (0, 3): The intersection of \( y = -3x + 3 \) and \( y = \frac{15-x}{5} \).
• (5, 2): The intersection of \( y = \frac{15-x}{5} \) and \( y = -2x + 12 \).

These points represent potential solutions to be evaluated in the objective function.

Linear Inequalities

Linear inequalities are expressions involving a linear function where one side is not equal to the other, signified by inequality symbols like \( \geq \) or \( \leq \). These inequalities are fundamental in defining constraints.

Each inequality translates into a half-plane on a coordinate graph. For example, \( y \geq 0 \) indicates a region above the x-axis. Similarly, inequalities like \( y \leq \frac{15-x}{5} \) guide how we shade a half-plane below this line.

In graphing these inequalities, you draw a straight line and shade the side signified by the inequality. Overlapping shaded areas from multiple inequalities form the feasible region. Each linear inequality plays a part in sculpting the boundaries of the possible solutions area.

Objective Function

The objective function is the formula you wish to optimize, usually through maximizing or minimizing a certain value. In our exercise, the objective function is \( C = 6x + y \).

This equation is calculated at each vertex of the feasible region. The goal is to find which vertex gives the lowest or highest value of \( C \), depending on whether we want to minimize or maximize. This happens because one of these corner points will typically be the optimal solution in linear programming.

In the problem solved, we calculated \( C \) at each relevant vertex:

• At (0, 3), \( C = 6(0) + 3 = 3 \)
• At (5, 2), \( C = 6(5) + 2 = 32 \)

Thus, the objective function achieves its minimum at the vertex (0, 3), highlighting one of the powerful uses of linear programming.