To determine the maximum value of the function Y=sqrt(3)x+sin(2x) on the interval [0, π], we need to first find the critical points of the function within that interval.

To find the critical points, we need to take the derivative of the function Y with respect to x and set it equal to zero. The derivative of sqrt(3)x is sqrt(3), and the derivative of sin(2x) is 2cos(2x). So, the derivative of Y is:

Y' = sqrt(3) + 2cos(2x).

Next, we set Y' equal to zero and solve for x:

sqrt(3) + 2cos(2x) = 0
2cos(2x) = -sqrt(3)
cos(2x) = -sqrt(3)/2
2x = 2pi/3, 4pi/3
x = pi/3, 2pi/3

So, the critical points within the interval [0, π] are x=pi/3 and x=2pi/3.

Next, we evaluate the function Y at these critical points and at the endpoints of the interval [0, π]:

Y(0) = sqrt(3)0 + sin(20) = 0
Y(pi/3) = sqrt(3)(pi/3) + sin(2(pi/3)) = pisqrt(3)/3 + sqrt(3)/2 ≈ 1.91
Y(2pi/3) = sqrt(3)(2pi/3) + sin(2(2pi/3)) = 2pisqrt(3)/3 - sqrt(3)/2 ≈ 1.91
Y(pi) = sqrt(3)pi + sin(2pi) = pi*sqrt(3) ≈ 5.44

Therefore, the maximum value of the function Y=sqrt(3)x+sin(2x) on the interval [0, π] is approximately 5.44, which occurs at x=pi.