Problem 15 - Finite Mathematics and Applied Calculus

Question

Sketch the region that corresponds to the given inequalities, say whether the region is bounded or unbounded, and find the coordinates of all corner points (if any). $$ \begin{aligned} 3 x+2 y & \geq 6 \\ 3 x-2 y & \leq 6 \\ x & \geq 0 \end{aligned} $$

Step-by-Step Solution

Answer

The region defined by the given inequalities is unbounded and extends infinitely upward along the \(x\)-axis. The corner point within this region is located at (2, 0).

Step 1: Graph the lines represented by the equalities

First, we need to graph the lines represented by the equalities: $$ \begin{aligned} 3x + 2y &= 6 \\ 3x - 2y &= 6 \\ \end{aligned} $$ To draw the lines, we can rewrite the equalities in slope-intercept form: $$ \begin{aligned} y &= -\frac{3}{2}x + 3\\ y &= \frac{3}{2}x - 3 \end{aligned} $$ Now, we can plot the lines on a coordinate plane.

Step 2: Identify the regions that satisfy the inequalities

Our original inequalities are: $$ \begin{aligned} 3x + 2y &\geq 6 \\ 3x - 2y &\leq 6 \\ x &\geq 0 \end{aligned} $$ To determine which regions satisfy each inequality, we can use the test point method to evaluate the inequality at a chosen test point. If the inequality holds true, the test point lies in the region defined by the inequality. Let's use the point (0,0) as a test point: $$ \begin{aligned} 3(0) + 2(0) &\ge 6 \Rightarrow &False \\ 3(0) - 2(0) &\le 6 \Rightarrow &True \\ 0 &\ge 0 \Rightarrow &True \end{aligned} $$ Since the point (0,0) satisfies the second and third inequalities, we can shade the region below the second line that satisfies these inequalities, including the \(x\)-axis (\(x \ge 0\)). The shaded region that does not satisfy the first inequality will be unbounded above.

Step 3: Determine if the region is bounded or unbounded

Based on our observation, we can identify that the region is unbounded, as it extends infinitely upward along the \(x\)-axis.

Step 4: Find the coordinates of any corner points

We will find the coordinates of the intersection points of the lines: $$ \begin{aligned} 3x + 2y &= 6 \\ 3x - 2y &= 6 \\ \end{aligned} $$ We can add the two equations to eliminate the \(y\)-variable: $$ (3x+2y)+(3x-2y) = 6+6 \\ 6x = 12 \\ x = 2 $$ Now, substitute \(x = 2\) back into either of the original equations to find \(y\): $$ 3(2) + 2y = 6 \\ 6 + 2y = 6 \\ 2y = 0 $$ So, \(y = 0\). Thus, the corner point is located at (2, 0). This is the only corner point within the defined region.