A question on differentiation, please help

Given that y=15x(3-x), calculate

a) the value of x when y is a maximum

b) the maximum value of y

a.) to solve for the maximum:

y = 45x - 15x²

y' = 45 - 30x

setting the slope to zero since, in a parabola,

the line tangent to the maximum is a straight line with

slope = 0;

0 = 45 - 30x

x = 3/2

b.) x = 3/2 when y is max.

y = 45 - 30 (3/2)

y = 0

solve y'=0 , then find y value