<s>[INST] Solution Verification: Find Extrema points of the function $f(x,y)=2^{3x+8y}$ that are on $x^2+y^2=1$. 
$f(x,y)=2^{3x+8y}$  
$x^2+y^2=1$

Polar Coordinates approach: 
Let $x=cos(t), y=sin(t), 0\le t \le 2\pi$ (We can see that this solved the circle equation). 
Substituting them into my function: 
$f = 2^{3\cos(t)+8\sin(t)} \Longrightarrow f=e^{(3\cos(t)+8\sin(t))ln(2)}$ 
Taking derivative to find critical points: 
$f' = \ln(2)(-3\sin(t)+8\cos(t))e^{(3\cos(t)+8\sin(t))ln(2)}=0$ 
Now $e^x>0$, so I need to find when 
$-3\sin(t)+8\cos(t)=0 \Longrightarrow 3 \sin(t)=8\cos(t) \Longrightarrow \tan(t) = \frac{8}{3} \Longrightarrow t=\arctan(\frac{8}{3}), \arctan(\frac{8}{3}) + \pi $
So I get two Points: 
$(\cos(\arctan(\frac{8}{3})) ,\sin(\arctan(\frac{8}{3})))$ 
$(\cos(\arctan(\frac{8}{3}) + \pi), \sin(\arctan(\frac{8}{3}))$ 
and now I need to find $f_{xx}, f_{yy}, f{xy}$ in order to know which point is minimum and which is maximum: 
$f_{x}=8^x3ln(2)$ 
$f_y = 256^y 8ln(2)$ 
$f_{xx} = 9\ln^2(2) 8^x$ 
$f_{yy} = 64ln^2(2) 256^y$ 
$f_{xy} = 0$ 
$f_{yx} = 0$.  

for first point: $f_{xx} = 8.97$, $f_{yy} = 5530.131$.  So $f_{xx}*f_{yy} - 0 > 0$ , and $f_{xx} > 0 $, which means it is a local minimum point. 

for second point: $f_{xx} = 2.03$ , $f_{yy} = 0.17$. and same as above, it is a minimum point. 
Would appreciate any help in approving this solution or finding mistakes in it, and if there's any more efficient approaches, thanks in advance.  
EDIT: Approach from Parcly Taxel answer: 
Checking maximum and minimum for $3x+8y$: 
$\phi (x,y,\lambda) = 3x+8y + \lambda(x^2+y^2-1)$ 
$\phi_x = 3 + 2\lambda x = 0 \Longrightarrow x= \frac{-3}{2\lambda}$ 
$\phi_y = 8 + 2 \lambda y = 0 \Longrightarrow y= \frac{-4}{\lambda}$  
$\phi_{\lambda} = x^2 +y^2 -1 = 0 \Longrightarrow \frac{9}{4\lambda^2} + \frac{16}{\lambda^2}-1=0 \Longrightarrow 9+64=4\lambda^2 \Longrightarrow \lambda^2 = \frac{73}{4} \Longrightarrow \lambda = \pm \frac{\sqrt{73}}{2}$

So: $x= \pm 0.3511 $ 
$y = \pm 0.9363$ 
$(0.3511, 0.9363) , (-0.3511, -0.9364)$
 [/INST] Since $\cos^2 t + \sin^2 t = 1$, can you maximise $3 \cos t + 8 \sin t$?
Without using calculus, you need $k (\sin(t+a)) = k(\sin t \cos \alpha + \cos t \sin \alpha)= 3 \cos t + 8 \sin t$, so $k \sin \alpha = 3, k \cos \alpha = 8$ which lets you find $\alpha$, then $k$.
This gives $\tan \alpha = \frac{8}{3}$ and since $\cos \arctan \alpha = \frac{1}{\sqrt{1+\alpha^2}}$, $k$ is $\frac{8}{1/\sqrt{1+(8/3)^2}} = \sqrt{73}$, hence the maximum value is $2^{\sqrt{73}}$.
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