Let y=(2√3+3)sinθ+2√3cosθ, and differentiate:
dy/dx=(2√3+3)cosθ-2√3sinθ=0 for maximum and minimum.
So (2√3+3)cosθ=2√3sinθ, tanθ=(2√3+3)/2√3=1+√3/2. Let sinθ=(2√3+3)/x and cosθ=2√3/x, where x2=(2√3+3)2+(2√3)2=12+12√3+9+12)=33+12√3.
Also, since tanθ=(-sinθ)/(-cosθ) as well as sinθ/cosθ, sinθ=-(2√3+3)/x and cosθ=-2√3/x.
y=±[(2√3+3)(2√3+3)/x+2√3(2√3)/x]=±[(2√3+3)2+(2√3)2]/x=±x2/x=±x.
x=7.333799 to 6 decimal places, but 2√3+√15=7.337085 to 6 decimal places.
The given maximum and minimum include the calculated maximum and minimum, but are overestimates by about 0.003286.