SOLUTION: Find the area of the region determined by the system y \ge |x|, \\ y \le -|2x + 1| + 6.

Question 1209886: Find the area of the region determined by the system
y \ge |x|, \\
y \le -|2x + 1| + 6.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's analyze the inequalities to determine the region.
1. **$y \ge |x|$**
* This represents the region above the V-shaped graph of $y = |x|$.
* $y = x$ for $x \ge 0$
* $y = -x$ for $x < 0$
2. **$y \le -|2x + 1| + 6$**
* This represents the region below the inverted V-shaped graph of $y = -|2x + 1| + 6$.
* To find the vertex of $y = -|2x + 1| + 6$, we set $2x + 1 = 0$, which gives $x = -\frac{1}{2}$.
* When $x = -\frac{1}{2}$, $y = -|0| + 6 = 6$. So, the vertex is $(-\frac{1}{2}, 6)$.
* For $2x + 1 \ge 0$, i.e., $x \ge -\frac{1}{2}$, $y = -(2x + 1) + 6 = -2x + 5$.
* For $2x + 1 < 0$, i.e., $x < -\frac{1}{2}$, $y = -(-2x - 1) + 6 = 2x + 7$.
Now, we need to find the intersection points of the graphs.
**Intersection of $y = |x|$ and $y = -|2x + 1| + 6$:**
* **Case 1: $x \ge 0$ and $x \ge -\frac{1}{2}$ (i.e., $x \ge 0$)**
* $x = -2x + 5$
* $3x = 5$
* $x = \frac{5}{3}$
* $y = \frac{5}{3}$
* Intersection point: $(\frac{5}{3}, \frac{5}{3})$
* **Case 2: $x < 0$ and $x \ge -\frac{1}{2}$ (i.e., $-\frac{1}{2} \le x < 0$)**
* $-x = -2x + 5$
* $x = 5$ (This is not in the interval, so no intersection)
* **Case 3: $x \ge 0$ and $x < -\frac{1}{2}$ (Impossible)**
* **Case 4: $x < 0$ and $x < -\frac{1}{2}$ (i.e., $x < -\frac{1}{2}$) **
* $-x = 2x + 7$
* $-3x = 7$
* $x = -\frac{7}{3}$
* $y = \frac{7}{3}$
* Intersection point: $(-\frac{7}{3}, \frac{7}{3})$
The intersection points are $(\frac{5}{3}, \frac{5}{3})$ and $(-\frac{7}{3}, \frac{7}{3})$.
Now, we need to find the area of the region.
We can split the area into two triangles.
* Triangle 1: Vertices $(-\frac{7}{3}, \frac{7}{3})$, $(-\frac{1}{2}, 6)$, and $(0, 0)$.
* Triangle 2: Vertices $(0, 0)$, $(-\frac{1}{2}, 6)$, and $(\frac{5}{3}, \frac{5}{3})$.
Area of Triangle 1:
Using the determinant formula:
$$ \frac{1}{2} \left| (-\frac{7}{3})(6 - 0) + (-\frac{1}{2})(0 - \frac{7}{3}) + 0(\frac{7}{3} - 6) \right| $$
$$ \frac{1}{2} \left| -14 + \frac{7}{6} \right| = \frac{1}{2} \left| \frac{-84 + 7}{6} \right| = \frac{1}{2} \left| \frac{-77}{6} \right| = \frac{77}{12} $$
Area of Triangle 2:
$$ \frac{1}{2} \left| 0(6 - \frac{5}{3}) + (-\frac{1}{2})(\frac{5}{3} - 0) + (\frac{5}{3})(0 - 6) \right| $$
$$ \frac{1}{2} \left| -\frac{5}{6} - 10 \right| = \frac{1}{2} \left| \frac{-5 - 60}{6} \right| = \frac{1}{2} \left| \frac{-65}{6} \right| = \frac{65}{12} $$
Total area:
$$ \frac{77}{12} + \frac{65}{12} = \frac{142}{12} = \frac{71}{6} $$
Final Answer: The final answer is $\boxed{\frac{71}{6}}$

RELATED QUESTIONS
Find the area of the region enclosed by y=2x^2 and... (answered by Cromlix)

Find the area of the region enclosed by the graph of x^2 + y^2 = 2x - 6y +... (answered by ikleyn)

Given that $-3 \le 7x + 2y \le 3$ and $-4 \le y - x \le 4$, what is the maximum possible... (answered by Alan3354)

HELPPP!!!! DUE TOMORROW!!: Given that $-3 \le 7x + 2y \le 3$ and $-4 \le y - x \le 4$,... (answered by richwmiller,ikleyn,MathTherapy,madyl,greenestamps)

Graph the feasible region for each system of inequalities 2x-y<1... (answered by stanbon)

Find the area of the region bounded by the x and y axes and the line y= 2x... (answered by sachi)

Find the area of the region bounded by the y-axis, x-axis and the line y=−2x+4. (answered by math_helper,MathLover1)

find the minumum value of f(x,y)=2x-y+2 for the polygonal convex set determined by this... (answered by RAY100)

Maximize:z=40x+12y Subject to: 6x-2y\le 9 3x+y\le 9 x\ge 0 y\ge... (answered by Theo)