Find the absolute max and min of f subject to the given constraint $f(x)=4x^4+y^4$ subject to the constraint $x^2+y^2=1$
My attempt at it was by finding partial derivitives:
$$f_x=16x^3 =0$$
$$f_y=4y^3=0$$
this means that we have an interior point $(0,0)$
Now Let $x=\cos(t)$, $y=\sin(t)$
$$f(\cos(t),\sin(t))=4\cos^4t+\sin^4t$$
$$<=> 3\cos^4t+1$$
we can now note that $f$ has max when $t=0,\pi$, or $2\pi $ and min when $t= \pi/t +k\pi)$
this means that we have the following potential points 
$f(1,0),f(-1,0), f(0,1), f(0,-1), f(0,0)$
$$f(0,0)=0$$
$$f(1,0)=4$$
$$f(0,1)=1$$
Max =4
Min = 0
Anyone can help verifying my answer, because i tried the lagrange way and it didnt work

Express $(x,y)$ in polar coordinates as $(\cos\theta,\sin\theta)$, and then let
$c= \cos^2 \theta$.  Note that  $\sin^2 \theta = 1-\cos^2 \theta$  so  $x^4=c^2$ and $y = (1-c)^2$.
Then $$4x^4+y^4 = 4c^2 + (1-c)^2 = 5c^2 -2c+1
$$
Since $c$, being $\cos^2\theta$, is restricted by $0 \leq |c|\leq 1$,
this is maximized at $c=1 = \cos^2\theta$, at which point $x=\pm1, y=0, f(x,y) =4 $.
Trivial differentiation locates the minimum at $c = 1/5$, at which point 
 $x=\pm \frac{\sqrt{5}}{5}, y=\pm 2\frac{\sqrt{5}}{5} , f(x,y) =\frac4{5} $.
