If the area of the region $$\{(x, y) : 0 \leq y \leq \min(2x, 6x - x^2)\}$$ is A, then 12A is equal to _____.

Intersection: $$2x = 6x - x^2 \implies x^2 - 4x = 0 \implies x = 0, 4$$.

Determine the "min" boundary:

o From $$x=0$$ to $$x=4$$, $$2x$$ is the line and $$6x-x^2$$ is the parabola.

o $$2x < 6x - x^2$$ when $$x < 4$$.

At $$x=1$$: $$2x=2, 6x-x^2=5 \rightarrow \min$$ is $$2x$$.

At $$x=5$$: $$2x=10, 6x-x^2=5 \rightarrow \min$$ is $$6x-x^2$$.

The curves intersect at $$x=4$$. For $$0 \le x \le 4$$, $$2x$$ is smaller. For $$x > 4$$, $$6x-x^2$$ is smaller until it hits the x-axis ($$y=0$$) at $$x=6$$.

Area $$A = \int_{0}^{4} 2x \, dx + \int_{4}^{6} (6x - x^2) \, dx$$

$$A = [x^2]_0^4 + [3x^2 - \frac{x^3}{3}]_4^6$$

$$A = (16 - 0) + [(3(36) - \frac{216}{3}) - (3(16) - \frac{64}{3})]$$

$$A = 16 + [(108 - 72) - (48 - 21.33)] = 16 + [36 - 26.66] =16 + \frac{28}{3} = \frac{76}{3}$$

$$12 \times \frac{76}{3} = 4 \times 76 = 304$$