Problem 4
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). $$ -x+2 y \geq 4 $$
Step-by-Step Solution
Verified Answer
The region corresponding to the given inequality is graphed by shading the area above the line \(y = \frac{1}{2}x + 2\). The region is unbounded and there are no corner points.
1Step 1: Graph the inequality
First, we need to graph the inequality. To do this, we can rewrite the inequality as an equation:
$$
-x + 2y = 4
$$
Now, we can solve for \(y\):
$$
y = \frac{1}{2}x + 2
$$
Next, we'll graph the line \(y = \frac{1}{2}x + 2\). The inequality states that the region of interest is above the line, so we'll shade the region above the line.
2Step 2: Determine if the region is bounded or unbounded
Now, let's determine if the region is bounded or unbounded. Since there's only one inequality, there aren't any other lines that constrain the region. Hence, the region is unbounded.
3Step 3: Identify corner points (if any)
Since the region is unbounded, there are no corner points for the region.
So, the final answer is:
- The region is graphed by shading the area above the line \(y = \frac{1}{2}x + 2\).
- The region is unbounded.
- There are no corner points.
Other exercises in this chapter
Problem 4
\(\begin{array}{cc}\text { Maximize } & p=x+2 y \\ \text { subject to } & x+y \leq 25 \\ y & \geq 10 \\ & 2 x-y \geq 0 \\ & x \geq 0, y \geq 0 .\end{array}\)
View solution Problem 4
\(\begin{array}{ll}\text { Maximize } & p=2 x+3 y \\ \text { subject to } & 3 x+8 y \leq 24 \\ & 6 x+4 y \leq 30 \\ & x \geq 0, y \geq 0 .\end{array}\)
View solution Problem 5
$$ \begin{array}{ll} \text { Maximize } & p=x+y+z+w \\ \text { subject to } & x+y+z \leq 3 \\ & y+z+w \leq 4 \\ & x+z+w \leq 5 \\ & x+y+w \leq 6 \\ & x \geq 0,
View solution Problem 5
Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. Maximize \(\begin{al
View solution