Problem 25 - 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). $$ \begin{array}{r} 2 x-y \geq 0 \\ x-3 y \leq 0 \\ x \geq 0, y \geq 0 \end{array} $$

Step-by-Step Solution

Answer

The region corresponding to the given inequalities is a bounded triangle with vertices at (0,0), (0,0), and (0,0), which, in this case, reduces to the single corner point at the origin (0,0).

Step 1: Rewrite inequalities as equalities

(Write the content here) We will rewrite the given inequalities as equalities to create the lines that define the region. \( \begin{cases} 2x-y= 0\\ x-3y= 0\\ x= 0, y= 0\\ \end{cases}\)

Step 2: Sketch the lines on a coordinate plane

Plot the lines obtained in step 1 on a coordinate plane. The lines are: \(y = 2x\) - a line with slope 2 and intercept (0, 0). \(y = \frac{1}{3}x\) - a line with slope 1/3 and intercept (0, 0). \(x = 0\) - the y-axis. \(y = 0\) - the x-axis.

Step 3: Shade the region that satisfies the inequalities

Now we need to determine which region in the plane satisfies all given inequalities: - For \(2x - y \geq 0\), choose a test point (1, 0). Since \(2(1) - 0 \geq 0\) is true, shade the region to the right of the line \(y = 2x\). - For \(x - 3y \leq 0\), choose a test point (0, 1). Since \(0 - 3(1) \leq 0\) is true, shade the region below the line \(y = \frac{1}{3}x\). - For \(x \geq 0\), shade all regions to the right of the y-axis. - For \(y \geq 0\), shade all regions above the x-axis. The region that satisfies all inequalities is the area in which all shaded regions overlap.

Step 4: Identify whether the region is bounded or unbounded

From the graph, it is evident that the region is bounded because it is enclosed on all sides, forming a triangle.

Step 5: Find coordinates of all corner points

There are three corner points in this region which are intersections of the lines: 1. Intersection of y-axis and the line \(y = 2x\): \(x=0\), \(2(0)-y=0\), so the point is (0, 0). 2. Intersection of x-axis and the line \(y= \frac{1}{3}x\): \(y=0\), \(\frac{1}{3}(0)-y=0\), so the point is (0, 0). This is the same as point 1. 3. Intersection of the lines \(y=2x\) and \(y= \frac{1}{3}x\): \(2x= \frac{1}{3}x\), so \(x=0\) and (in both equations) \(y=0\). Therefore, corner point is (0, 0). In this specific case, there is only one corner point: (0, 0).