Maximal and minimal value of a function $4x^2-4xy+y^2$ at the border $x^2+y^2=25$. So my first thought was to find the critical points of the $4x^2-4xy+y^2$ and I calculated.
$$F_{x}=8x-4y,\quad F_{y}=2y-4x$$
and I found out that critical point are satisfying $x=2y$.
Then I put then to equation $x^2+y^2=25$.
And I calculated that $x=\sqrt5,y=2\sqrt5$, and $x=-\sqrt5$, $y=-2\sqrt5$. 
But I don't know if they are maximum or minimum.
My last idea was to to put $x^2=25-y^2 $ to the function $4x^2-4xy+y^2$ and then calculate the derivative. But I do not know if it is a good solution.

$x =u + 2v\\
y = 2u - v$
$4x^2 -4xy + y^2 = 4(u^2 + 4uv + 4v^2) - 4(2u^2 -3uv - 2v^2) + (4u^2 - 4uv + v^2) = 25 v^2$
$x^2 + y^2 = (u^2 + 4uv + 4v^2) + (4u^2 - 4uv + v^2) = 5u^2 + 5v^2 = 25$ 
Minimize $25v^2$ constrained by: $5u^2 + 5v^2 = 25$
$v = 0, u = \pm \sqrt 5\\
(x,y) = (\sqrt 5,2\sqrt 5), (-\sqrt 5, -2\sqrt5)$
Maximize $25v^2$ constrained by: $5u^2 + 5v^2 = 25$
$v = \pm \sqrt 5, u=  0\\
(x,y) = (2\sqrt 5, -\sqrt 5), (-2\sqrt 5, \sqrt 5)\\
4x^2 -4xy +y^2 = 125$
