I have the following problem: $$Min/Max \quad x_1^2 - 4x_1x_2 + 4x_2^2 - 12x_1 - 18x_2 $$ with the constraint $(x_1 - 2)^2 + 3(x_2 - 1)^2 = 7$. For this problem, I'm also working with this form of the constraint: $x_1^2 - 4x_1 + 3x_2^2 - 6x_2 = 0$.
After making sure the conditions for the existence of a maximum/minimum are met, I have a couple of notes: I've tried solving this without a substitution with the Lagrangian being: $L: \mathbb{R}^2 \times \mathbb{R} \rightarrow \mathbb{R}, L(x,u) = x_1^2 - 4x_1x_2 + 4x_2^2 - 12x_1 - 18x_2 + u (x_1^2 - 4x_1 + 3x_2^2 - 6x_2)$
Then, trying to find stationary points, I did the partial derivatives:
$\dfrac{\partial L(x_1, x_2, u)}{\partial x_1} := 2x_1 - 4x_2 - 12 + 2x_1u - 4u = 0$
$\dfrac{\partial L(x_1, x_2, u)}{\partial x_2} := -4x_1 + 8x_2 - 18 + 6x_2u - 6u = 0 $
$\dfrac{\partial L(x_1, x_2, u)}{\partial u} := x_1^2 - 4x_1 + 3x_2^2 - 6x_2 = 0 $
Trying to solve for a complete set of feasible $u$'s turned out to be impossible without the use of software, which is not allowed. I have to solve this on paper.
My supervisor suggested using the following substitution: $$f_0 = x_1^2 - 4x_1x_2 + 4x_2^2 - 12x_1 - 18x_2 = (x_1 - 2x_2)^2 - 6(2x_1 + 3x_2) $$ $$z_1 = x_1 - 2x_2$$ $$z_2 = 2x_1 + 3x_2$$ $$ \Rightarrow f_0 = z_1^2 - 6z_2^2$$
With this substitution, the Lagrangian should then apparently be separable.
Please, can you help me incorporate the sub. into this problem, and show me how the Lagrangian is going to separable? I've tried transforming the constraint with the $z$ sub. but got a ghastly looking thing that surely doesn't look better than the original. I don't know what I'm missing here.
Thanks a lot