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.
 A: It's obvious that the minimum is $0$.
For $x=2\sqrt5$ and $y=-\sqrt5$ we get a value $125$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$(2x-y)^2\leq5(x^2+y^2)$$ or
$$(x+2y)^2\geq0.$$
Done!
A: Hint. Note that $F(x,y)=4x^2-4xy+y^2=(2x-y)^2\geq 0$.  
Note that the level curves are the lines parallel to $y=2x$. 
Therefore the minimal value is $0$ and it is attained at the intersections of line $y=2x$ and the circle $x^2+y^2=5^2$.
On the other hand, the maximum value is attained when the level curve is tangent to the circle that is along the line $y=-x/2$ (which is orthogonal to $y=2x$).
A: Using $F_x$ and $F_y$ as you have done is not going to help you complete the problem. Because along the constraint curve there can be extrema where $F_x$ and $F_y$ are nonzero. 
In the comments, you said you do not know the method of Lagrange multipliers. So instead, try this. The points along the constraint curve are actually points on a circle $(5\cos(t),5\sin(t))$. So sub in $x=5\cos(t)$ and $y=5\sin(t)$ into your formula for $F$, and then analyze it as a one-variable function of $t$. When you find critical $t$-values, substitute them back to find the corresponding $x$ and $y$.
A: $x =u + 2v\\
y = 2u - v$
$4x^2 -4xy + y^2 = 4(u^2 + 4uv + 4v^2) - 4(2u^2 -3uv - 2v^2) + (4u^2 - 4uv + v^2) = 25 v^2$
$x^2 + y^2 = (u^2 + 4uv + 4v^2) + (4u^2 - 4uv + v^2) = 5u^2 + 5v^2 = 25$ 
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: $5u^2 + 5v^2 = 25$
$v = \pm \sqrt 5, u=  0\\
(x,y) = (2\sqrt 5, -\sqrt 5), (-2\sqrt 5, \sqrt 5)\\
4x^2 -4xy +y^2 = 125$
A: Maybe this way is a bit simpler: by Weierstrass theorem, that functions attains a global maximum and a global minimum in the compact $x^2+y^2=25$. So you might as well just compare the images of those two points; the greater will be the maximum, and the lesser will be the minimum. 
