Let the area of the region $\{(x,y): 2y\leq x^2+3,\; y+|x|\leq 3,\; y\geq|x-1|\}$ be $A$. Then $6A$ is equal to:
Step-by-Step Solution
Key Concept: Sketch all three boundary curves, find the vertices of the feasible polygon, then compute the area as (area of enclosing rectangle) minus (area of excluded triangular regions).
Intersections yield vertices at $A(1,0)$, $B(2,1)$, $C(1,2)$, $D(0,3)$, $E(-1,2)$.
$A = 4 - 2\int_0^1\!\left[(3-x)-\frac{x^2+3}{2}\right]dx = 4 - 2\int_0^1\!\left(3-x-\frac{x^2}{2}-\frac{3}{2}\right)dx$
$= 4 - 2\left[\frac{3}{2}x - \frac{x^2}{2} - \frac{x^3}{6}\right]_0^1 = 4 - 2\left(\frac{3}{2}-\frac{1}{2}-\frac{1}{6}\right) = 4 - \frac{5}{3} = \frac{7}{3}$.
$6A = 14$.
Correct Answer: 3