Maximizing $ 3^{\sin^2{\theta}} \cdot 27^{\cos^2{\theta}} + 8^{\sin{\theta}}\cdot 16^{\cos{\theta}} $ 
Find the maximum value of $$ 3^{\sin^2{\theta}} \cdot 27^{\cos^2{\theta}} + 8^{\sin{\theta}}\cdot 16^{\cos{\theta}} $$
Where does maximum of the expression occur?

I can find maxima of individual terms easily, but since they occur at different values of $\theta$, that is not getting anywhere.
If I differentiate, it becomes very tedious.

Is there any clever rearrangement or logic, without calculus, which will get me max value of this expression?

I tried arithmetic mean greater than geometric mean but could not get anywhere.
Thank you.
 A: You can rewrite the expression as
$$y=3^{\cos (2 t)+2}+2^{3 \sin (t)+4 \cos (t)}$$
The first piece will reach a  maximum value $(3^3=27)$ when $t=0$ and the second will reach a  maximum value $(2^5=32)$ when $t=2 \cot ^{-1}(3)$. This means that the maximum value of $y$ will not exceed $27+32=59$; on the other side, the maximum value of $y$ will be larger than $3^3+2^4=43$.
Plotting the function, you could see that the maximum occurs "not very far" of $t=0$. Using composition of Taylor series (this is quite tedious),
$$y=43+48  \log (2)\,t+ \left(72 \log ^2(2)-32 \log (2)-54 \log (3)\right)t^2+
   O\left(t^3\right)$$
Computing the derivative and making it equal to zero gives
$$t_{max}=-\frac{12 \log (2)}{36 \log ^2(2)-16 \log (2)-27 \log (3)}\approx 0.354603$$ which gives $y_{max}\approx 48.4354$.
A numerical optimization would lead to $t_{max}\approx 0.330073$ and  $y_{max}\approx 48.4634$. 
For sure, you could use Newton method to find the zero of
$$y'=\log (2) 2^{3 \sin (t)+4 \cos (t)} (3 \cos (t)-4 \sin (t))-2 \log (3) \sin (2 t)
   3^{\cos (2 t)+2}$$ and the iterates would be
$$\left(
\begin{array}{cc}
 n & t_n \\
 0 & 0.000000 \\
 1 & 0.354603 \\
 2 & 0.329806 \\
 3 & 0.330073
\end{array}
\right)$$
Finally, being very lazy, using a small programmable pocket calculator set in degrees, using steps of $1 ^{\circ}$, the maximum of $y$ was reached for $t=19 ^{\circ}=0.331613$ radians and found to be $48.4633$.
A: Write your term as $$3^{3\cos^2(\theta)+\sin^2(\theta)}+2^{3\sin(\theta)+4\cos(\theta)}$$
