# Given the function y= 3 sin [2 (x- 30 degrees)] , find the smallest positive value for x that gives a maximum value for y.

### 1 Answer | Add Yours

`y=3 sin(2(x-30))`

for maximum value of y we want derivative of y with respect to x.

`(dy)/(dx)=6cos(2(x-30))`

for maximum

`(dy)/(dx)=0`

`6cos(2(x-30))=0`

`Cos(2(x-30))=Cos(90)`

`2(x-30)=90`

`2x-60=90`

`2x=150`

`x=75^o`

`(d^2y)/(dx^2)=-12sin(2(x-30))`

`(d^2y)/(dx^2)}_{x=75^o}=-12 <0`

Thus x=75 degree then y has aximum value.

**Thus answer to question is x=75 degree.**