The area bounded by 0 ≤ y ≤ min{2x,6x − x2 } and x-axis is A. Then 12A is

The correct answer is 304.

We are given the region bounded by:
$0\le y\le min\{2x,6x-x^2\}$
and the $x$-axis (which is $y=0$).

To find the boundary of the region, we first find the intersection points of the line $y=2x$ and the parabola $y=6x-x^2$ by equating them:
$2x=6x-x^2$
$x^2-4x=0$
$x(x-4)=0$
So, the curves intersect at $x=0$ and $x=4$.

Now, let's analyze the behavior of the two functions for $x\ge 0$:
1. For $0\le x\le 4$, we have $2x\le 6x-x^2$. Thus, the minimum of the two functions is $y=2x$.
2. For $x\ge 4$, we have $6x-x^2\le 2x$. Thus, the minimum of the two functions is $y=6x-x^2$.
3. Since we require $y\ge 0$, the boundary curve $y=6x-x^2$ meets the $x$-axis ($y=0$) at $x=6$ (since $6x-x^2=0\Rightarrow x=6$ for $x>0$).

The graph illustrating this region is shown below:

Therefore, the area $A$ is the sum of two integrals:
$A={\int }_0^{4}2xdx+{\int }_4^6(6x-x^2)dx$

Let us compute the first integral:
${\int }_0^{4}2xdx={\left[x^2\right]}_0^4=16-0=16$

Now let us compute the second integral:
${\int }_4^6(6x-x^2)dx={\left[3x^2-\frac{x^3}{3}\right]}_4^6$

Evaluating at the upper limit $x=6$:
$3\left(36\right)-\frac{216}{3}=108-72=36$

Evaluating at the lower limit $x=4$:
$3\left(16\right)-\frac{64}{3}=48-\frac{64}{3}=\frac{144-64}{3}=\frac{80}{3}$

Subtracting the two values:
$36-\frac{80}{3}=\frac{108-80}{3}=\frac{28}{3}$

Adding the two parts of the area:
$A=16+\frac{28}{3}=\frac{48+28}{3}=\frac{76}{3}$

We are required to find $12A$:
$12A=12\times \frac{76}{3}=4\times 76=304$