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Chapter 4: Problem 3

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). $$ -x-2 y \leq 8 $$

Short Answer

Expert verified
The boundary line of the given inequality \(-x - 2y \leq 8\) is \(y = -\frac{1}{2}x - 4\). The region corresponding to the inequality lies below the line, as determined by the test point (0, 0). The region is unbounded and has no corner points.

Step by step solution

01

Rewrite the inequality as an equation

Rewrite the inequality \(-x - 2y \leq 8\) as an equation to find the boundary line: \[ -x - 2y = 8 \]
02

Solve for y to find the line in slope-intercept form

Solve for y in the equation \(-x - 2y = 8\) Add x to both sides: \[ -2y = x + 8 \] Divide both sides by -2: \[ y = -\frac{1}{2}x - 4 \]
03

Determine the region corresponding to the inequality

Now, we need to find the region that corresponds to the original inequality \(-x - 2y \leq 8\). Let's use a test point to determine if the solution lies above or below the line \(y = -\frac{1}{2}x - 4\). A good point to choose is the origin (0, 0). Substitute x = 0 and y = 0 into the inequality: \[ - (0) - 2(0) \leq 8 \] Simplify: \[ 0 \leq 8 \] Since this statement is true, we know that the origin (0, 0) is in the solution region. This means that the correct region is below the line \(y = -\frac{1}{2}x - 4\).
04

Sketch the region and determine if it is bounded or unbounded

To sketch the region, begin by plotting the line \(y = -\frac{1}{2}x - 4\). Then, shade the region below the line, which contains the origin (0, 0). Upon examining the graph, we can see that the region is unbounded. It extends infinitely downward and to the right.
05

Identify corner points

Since the region is unbounded, there are no corner points. The region extends infinitely in two directions, so there are no finite points where two boundaries meet. In conclusion, the region corresponding to the inequality \(-x - 2y \leq 8\) is unbounded and has no corner points.

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