Problem 18 - Algebra 2

Question

Graph each system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the given function for this region. $$ \begin{array}{l}{y \geq 1} \\ {2 \leq x \leq 4} \\ {x-2 y \geq-4} \\ {f(x, y)=3 y+x}\end{array} $$

Step-by-Step Solution

Answer

Vertices: (2,1), (4,1), (2,3), (4,4). Max: 16 at (4,4). Min: 5 at (2,1).

Step 1: Graph the Inequality y ≥ 1

The inequality \( y \geq 1 \) implies that the region of interest includes all points where the \( y \)-coordinate is greater than or equal to 1. To represent this, draw a horizontal line across the graph at \( y = 1 \). Shade the region above and including this line to indicate that these are the permissible values for \( y \).

Step 2: Graph the Inequality 2 ≤ x ≤ 4

This compound inequality tells us that \( x \) must be between 2 and 4, inclusive. Draw two vertical lines, one at \( x = 2 \) and another at \( x = 4 \). Shade the region between these two lines as \( x \) can take any value from 2 to 4.

Step 3: Graph the Inequality x - 2y ≥ -4

To graph \( x - 2y \geq -4 \), first rewrite it in slope-intercept form: \( y \leq \frac{x + 4}{2} \). This is a line with a slope of \( \frac{1}{2} \) and a y-intercept of 2. Draw this line, and shade the region below and including the line. This represents the inequality \( x - 2y \geq -4 \).

Step 4: Identify the Feasible Region

The feasible region is the overlap of the shaded areas from Steps 1, 2, and 3. This region is bounded by the lines \( y = 1 \), \( x = 2 \), \( x = 4 \), and the line \( x - 2y = -4 \). Identify the vertices of this region by finding the points of intersection between the boundary lines of the inequalities.

Step 5: Find the Coordinates of the Vertices

The intersections to consider are: \((x = 2, y = 1)\), \((x = 4, y = 1)\), \((x = 2, x - 2y = -4)\), and \((x = 4, x - 2y = -4)\). Calculate the intersections:1. \( y = 1 \) and \( x = 2 \) gives \( (2, 1) \).2. \( y = 1 \) and \( x = 4 \) gives \( (4, 1) \).3. \( x = 2 \), substitute into \( x - 2y = -4 \) to get \( 2 - 2y = -4 \rightarrow y = 3 \), giving \( (2, 3) \).4. \( x = 4 \), substitute into \( x - 2y = -4 \) to get \( 4 - 2y = -4 \rightarrow y = 4 \), giving \( (4, 4) \).So, the vertices are \((2, 1), (4, 1), (2, 3), (4, 4)\).

Step 6: Find Maximum and Minimum Values of f(x, y) = 3y + x

Evaluate the function \( f(x, y) = 3y + x \) at each of the vertices.1. At \((2, 1)\), \( f(2, 1) = 3 \times 1 + 2 = 5 \).2. At \((4, 1)\), \( f(4, 1) = 3 \times 1 + 4 = 7 \).3. At \((2, 3)\), \( f(2, 3) = 3 \times 3 + 2 = 11 \).4. At \((4, 4)\), \( f(4, 4) = 3 \times 4 + 4 = 16 \).The maximum value is 16 at \((4, 4)\) and the minimum value is 5 at \((2, 1)\).

Key Concepts

Feasible Region

In the realm of systems of inequalities, the feasible region plays an essential role. It represents the set of all possible solutions that satisfy each inequality in the system. When graphing, it is the area where all shaded regions from individual inequalities overlap.

For our exercise, we identified the feasible region by intersecting the shaded areas resulting from graphing the inequalities:

• \( y \geq 1 \) — indicating all points above the horizontal line at \( y = 1 \).
• \( 2 \leq x \leq 4 \) — representing the vertical strip between lines \( x = 2 \) and \( x = 4 \).
• \( x - 2y \geq -4 \) — encompassing the area below and including the line \( y \leq \frac{x+4}{2} \).

Together, these constraints bound a specific area on the graph that is feasible for solutions. Understanding this overlap ensures that every point within it meets all the conditions of the given inequalities.

Vertices of Inequalities

Vertices are crucial in analyzing the feasible region. These points occur at the intersections of the boundary lines of the inequalities. They provide necessary reference points for determining extremum values of functions within the region.

In this problem, we found the vertices by calculating the points where the boundary lines intersect:

• Intersection of \( y = 1 \) and \( x = 2 \) yields \((2,1)\).
• Intersection of \( y = 1 \) and \( x = 4 \) yields \((4,1)\).
• Intersection of \( x = 2 \) with the line \( x - 2y = -4 \) gives \((2,3)\).
• Intersection of \( x = 4 \) with the line \( x - 2y = -4 \) gives \((4,4)\).

These vertices \((2, 1), (4, 1), (2, 3), (4, 4)\) define the corners of the feasible region. They are essentially the backbone for analyzing further properties like maximum or minimum values of given functions within the region.

Minimum and Maximum Values

Once you've identified the vertices of the feasible region, you can determine the minimum and maximum values of a function over this region. This process involves evaluating the function at each vertex and comparing the results.

For the function \( f(x, y) = 3y + x \), we calculated the values at each of the vertices:

• At \((2, 1)\), the function's value is 5.
• At \((4, 1)\), the function's value is 7.
• At \((2, 3)\), the function's value is 11.
• At \((4, 4)\), the function's value stands at 16.

With these values in mind, the maximum value of \( f(x, y) \) is found to be 16 at the vertex \((4, 4)\). The minimum value is 5 at the vertex \((2, 1)\). Evaluating the function in this way provides a clear indication of the extremum points within the feasible region.

Graphing Inequalities

Graphing inequalities might sound complex, but it's a straightforward process once you know what each part of the inequality represents. Every inequality has a boundary line, which you either include or exclude based on whether the inequality is 'less than or equal to' or 'greater than or equal to'.

Here is how we approached graphing the given inequalities:

• For \( y \geq 1 \), we drew a horizontal line at \( y = 1 \), shading above this line.
• The inequality \( 2 \leq x \leq 4 \) led to two vertical lines at \( x = 2 \) and \( x = 4 \), shading the strip between them.
• The last inequality \( x - 2y \geq -4 \) or \( y \leq \frac{x+4}{2} \) required converting to a slope-intercept form line, which we drew and shaded below.

Ensure the areas of overlap reflect all solution sets, representing where every condition is concurrently satisfied. This approach makes handling inequalities logical and systematic, reducing any confusion about which side of a boundary line to shade.