Since the function was apparently altered after at least some of the answers were posted, I'll address a peculiarity of this problem that may have raised **Ilda**'s original concern.

It should be remarked upon that while the "level curves" of the function $ \ f(x, \ y) \ = \ x^2 \ + \ 4xy \ + \ 4y^2 \ $ may give the impression of being (rotated) ellipses, this is in fact the equation of a *degenerate* conic: since $ \ x^2 \ + \ 4xy \ + \ 4y^2 \ = \ (x \ + \ 2y)^2 \ $ , a level curve is actually a pair of *parallel lines*. Moreover, since we are dealing only with real-valued coordinates, the **minimum value** of the function is **zero**, which "collapses" that level curve to a single line through the origin. This is the case quite *independently* of the constraint.

So a geometric interpretation of the Lagrange-multiplier problem is to find the value(s) of the function for which the degenerate conic is tangent to the "constraint ellipse" $ \ x^2 \ + \ 2y^2 \ = \ 4 \ $ . For simple convenience, and as a reminder of the possible values of the function, we will express the level curves of $ \ f(x, \ y) \ $ as $ \ (x \ + \ 2y)^2 \ = \ c^2 \ $ . Here is a graph of the situation:

Now, we *can* solve some of this problem without calculus, just for the purpose of investigating the nature of the extremization. If we describe the parallel lines by $ \ x \ + \ 2y \ = \ \pm c \ $ , we can locate any possible intersection points with the constraint ellipse from

$$ \ ( \ \pm c \ - \ 2y \ )^2 \ + \ 2y^2 \ = \ 4 \ \ \Rightarrow \ \ 6y^2 \ \mp \ 4cy \ + \ (c^2 \ - \ 4) \ = \ 0 \ \ . $$

The discriminant of this quadratic equation is $ \ ( \ \mp 4c \ )^2 \ - \ 4 \cdot 6 \ (c^2 \ - \ 4) \ = \ 96 \ - \ 8c^2 \ $ . Each line has a single intersection with the ellipse when this discriminant is zero, which occurs for $ \ c^2 \ = \ 12 \ $ . The discriminant is negative for larger values of $ \ c^2 \ $ , meaning that there would be **no** intersection points, so the maximum value of $ \ f(x, \ y) \ $ under the constraint is $ \ 12 \ $ . As we said above, the minimum value is necessarily zero, which does produce intersection points with the ellipse; these, however, are *not* tangent points.

[**Side note --** with a little further work, we can determine the two tangent points for $ \ f(x, \ y) \ = \ 12 \ $ to be $ \ \left( \pm \frac{2}{3} \sqrt{3} , \ \pm \frac{2}{3} \sqrt{3} \right) \ $ , which are placed symmetrically about the origin, as we expect when both the function to be extremized *and* the constraint function are symmetrical under inversion (what is also called "symmetry about the origin"). Ordinarily in an optimization of this sort, as **joriki** observes, each of these points would be associated with the maximum *or* minimum of the function. We see something of that here if we write the equations of the parallel lines individually, but in our situation, those lines represent only *one* value of the function.]

We can now look at what the Lagrange-multiplier method leads us to. The Lagrange equations for (the final version of) the function under the constraint are

$$ 2x \ + \ 4y \ = \ \lambda \cdot 2x \ \ , \ \ 4x \ + \ 8y \ = \ \lambda \cdot 4y \quad \mathbf{[1]} $$

$$ \Rightarrow \ \ ( \ 1 \ - \ \lambda \ ) \ x \ + \ 2y \ = \ 0 \ \ , \ \ x \ + \ ( \ 2 \ - \ \lambda \ ) \ y \ = \ 0 \ \ . \quad \mathbf{[2]} $$

[this differs from **Christian Blatter**'s answer, owing to the subsequent revision of the posted function]

We could also solve Equations **[ 1 ]** *for*

$$ \lambda \ = \ \frac{x \ + \ 2y}{x} \ = \ \frac{x \ + \ 2y}{y} \ \ . $$

For $ \ x \ \ne \ 0 \ , \ y \ \ne \ 0 \ $ , we may write

$$ x^2 \ + \ 2xy \ = \ xy \ + \ 2y^2 \ \ \Rightarrow \ \ x^2 \ + \ xy \ - \ 2y^2 \ = \ 0 \ \ ; $$

we can put this together with the fact that (since there is symmetry about the origin for $ \ f(x, \ y) \ $ and the constraint function) the extrema lie on lines through the origin $ \ y \ = \ mx \ $ to obtain

$$ x^2 \ + \ mx^2 \ - \ 2 \ m^2 x^2 \ = \ 0 \ \ \Rightarrow \ \ x^2 \ ( 2m^2 \ - \ m \ - \ 1 ) \ = \ 0 $$

$$ \Rightarrow \ \ m \ = \ 1 \ \ , \ \ -\frac{1}{2} \ \ \text{for} \ \ x \ \ne \ 0 \ \ . $$

From these slopes, we can determine

$$ \mathbf{m = 1 :} \quad y \ = \ x \ \ \Rightarrow \ \ x^2 \ + \ 2x^2 \ = \ 4 \ \ \Rightarrow \ \ x^2 \ = \ \frac{4}{3} \ \ , $$
$$ f \left(\frac{4}{3} , \ \frac{4}{3} \right) \ = \ x^2 \ + \ 4x^2 \ + \ 4x^2 \ = \ 9 \ \cdot \frac{4}{3} \ = \ 12 \ ; \ \text{also} \ \ \lambda \ = \ \frac{x \ + \ 2x}{x} \ = \ 3 \ \ ; $$

$$ \mathbf{m = -\frac{1}{2} :} \quad x \ + \ 2y \ = \ 0 \ \ , \ \ f \left(x , \ -\frac{1}{2}x \right) \ = \ x^2 \ - \ 2x^2 \ + \ x^2 \ = \ 0 \ \ ; $$
$$ \lambda \ = \ \frac{x \ + \ 2y}{x} \ = \ \frac{x \ + \ 2y}{y} \ = \ 0 \ \ , \ \ \text{for} \ \ x, \ y \ \ne \ 0 \ \ . $$

We obtain from this the extremal values of our function, and also show that the minimum value corresponds to $ \ \lambda \ = \ 0 \ $ , which does not have tangent points on the constraint ellipse.

We can arrive at this latter result in another way: Equations **[ 2 ]** become a linearly dependent system for $ \ \lambda \ = \ 0 \ $ , producing the single equation $ \ x \ + \ 2y \ = \ 0 \ $ with the same conclusions. The features we have seen here are not typical of an optimization problem involving functions related to *non-degenerate* conic sections.