Problem 24 - 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). $$ \begin{aligned} -3 x+2 y & \leq 5 \\ 3 x-2 y & \geq 6 \\ y & \leq x / 2 \\ x \geq 0, y \geq & 0 \end{aligned} $$

Step-by-Step Solution

Answer

The region corresponding to the given inequalities is bounded and lies in the first quadrant. The corner points of the region are \(P_1=(0,0)\), \(P_2=(2,1)\), and \(P_3=(4,5)\).

Step 1: Rewrite inequalities as equalities

Rewrite the inequalities as equalities to graph them: $$ \begin{aligned} -3x + 2y &= 5 \\ 3x - 2y &= 6 \\ y &= \frac{x}{2} \end{aligned} $$

Step 2: Sketch the individual lines

To graph the lines, first rewrite equations in slope-intercept form (y = mx + b) where possible, then plot the lines: $$ \begin{aligned} y &= \frac{3}{2}x + \frac{5}{2} \\ y &= \frac{3}{2}x - 3 \\ y &= \frac{x}{2} \end{aligned} $$ Also, note the two conditions \( x \geq 0 \) and \( y \geq 0 \). This means that the region lies in the first quadrant.

Step 3: Identify the region formed by the inequalities

Now that we have the lines plotted, we need to fill in the regions determined by the inequalities. Recall the inequalities and their corresponding lines: $$ \begin{aligned} -3 x+2 y & \leq 5 & (1) \ \ y &= \frac{3}{2}x + \frac{5}{2} \\ 3 x-2 y & \geq 6 & (2) \ \ y &= \frac{3}{2}x - 3 \\ y & \leq x / 2 & (3) \ \ y &= \frac{x}{2} \\ x \geq 0, y \geq & 0 & (4) \end{aligned} $$ (1) Keep the region below or on the line. (2) Keep the region above or on the line. (3) Keep the region below or on the line. (4) Keep the region in the first quadrant.

Step 4: Check if the region is bounded or unbounded

A region is bounded when it's contained within a finite area. Looking at our graph, the region formed by the intersections of the lines and inequalities is enclosed, so the region is bounded.

Step 5: Find the corner points of the region

Corner points are the points of intersection of the boundary lines. We have 3 corner points: 1. Intersection of lines (1), (3), and (4): \( P_1 = (0, 0) \) 2. Intersection of lines (1) and (3): Solving the system of equations, $$ \begin{cases} y = \frac{3}{2}x + \frac{5}{2} \\ y = \frac{x}{2} \end{cases} $$ We find that \( P_2 = (2,1) \). 3. Intersection of lines (1) and (2): Solving the system of equations, $$ \begin{cases} y = \frac{3}{2}x + \frac{5}{2} \\ y = \frac{3}{2}x - 3 \end{cases} $$ We find that \( P_3 = (4,5) \). So, the corner points are \(P_1=(0,0)\), \(P_2=(2,1)\), and \(P_3=(4,5)\).