Problem 3
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). $$ -x-2 y \leq 8 $$
Step-by-Step Solution
Verified Answer
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.
1Step 1: Rewrite the inequality as an equation
Rewrite the inequality \(-x - 2y \leq 8\) as an equation to find the boundary line:
\[
-x - 2y = 8
\]
2Step 2: 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:
\[
y = -\frac{1}{2}x - 4
\]
3Step 3: 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 \(y = -\frac{1}{2}x - 4\).
4Step 4: 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.
5Step 5: 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.
Other exercises in this chapter
Problem 3
\(\begin{array}{lc}\text { Maximize } & p=12 x+10 y \\ \text { subject to } & x+y \leq 25 \\ & x \quad \geq 10 \\ & -x+2 y \geq 0 \\ x & \geq 0, y \geq 0 .\end{
View solution Problem 3
\(\begin{array}{ll}\text { Maximize } & p=x-y \\ \text { subject to } & 5 x-5 y \leq 20 \\ & 2 x-10 y \leq 40 \\ & x \geq 0, y \geq 0 .\end{array}\)
View solution Problem 4
$$ \begin{array}{ll} \text { Minimize } & c=2 s+2 t+3 u \\ \text { subject to } & s \quad+u \geq 100 \\ & 2 s+t \quad \geq 50 \\ & s \geq 0, t \geq 0, u \geq 0
View solution Problem 4
Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. \(\begin{aligned} \t
View solution