Maximizing linear objective subject to quadratic equality constraint

Question

I've been trying to get through some practice questions on the Karush-Kuhn-Tucker (KKT) theorem but I can't seem to answer the following.

Given $f, g : \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x) := x_1 + x_2$ and $g(x) := x_1^2+3x_1x_2+3x_2^2-3$, respectively, $$\begin{array}{ll} \underset{x \in \mathbb{R}^n}{\text{maximize}} & f(x)\\ \text{subject to} & g(x) = 0\end{array}$$

My attempt:

$$\nabla f(x)=\begin{pmatrix} 1\\ 1\\ \end{pmatrix}$$ $$\nabla g(x)=\begin{pmatrix} 2x_1+3x_2\\ 3x_1+6x_2\\ \end{pmatrix}$$ and by complementary slackness $\lambda[x_1^2+3x_1x_2+3x_2^2-3]=0$ and $\lambda\geq0$

By first order conditions, I get $\lambda[2x_1+3x_2]=1$ and $\lambda[3x_1+6x_2]=1$

I checked WolframAlpha and the answer should be (3,-1) but I can't seem to figure out the right steps to solve this optimization.

What is your question? There is no $h$ given. There is no slack with an equality constraint. — copper.hat, Dec 14 '21 at 21:32
@copper.hat I'm pretty much stuck and can't solve this optimization. Also, I don't know why but I'm getting confused on when to use the complementary slack constraint. I looked it up and most solutions would make use of the constraint. — ji zhi, Dec 14 '21 at 21:40
This is Lagrange multipliers, there is no inequality constraint. You can solve the equations you got from the first order conditions to get a relationship between $x_1, x_2$. Then use this relationship with $g(x) = 0$ to obtain solutions and figure out which one gives the lower value of $f$. — copper.hat, Dec 14 '21 at 21:43
Note that $g'(x)$ is an invertible matrix times $x$. You can solve explicitly for $x$ in terms of $\lambda$. — copper.hat, Dec 14 '21 at 21:56
I honestly don't follow. I tried equating just to get an expression and plug it back into the constraint, but I don't think I'm doing it right as I'm not getting the right values. — ji zhi, Dec 14 '21 at 22:05
Why don't you read my second comment carefully and think about it a little. — copper.hat, Dec 14 '21 at 22:25

score 1 · Answer 1 · answered Dec 16 '21 at 19:54

You were so close. Solving the linear system $\lambda[2x_1+3x_2]=1,\lambda[3x_1+6x_2]=1$ yields $x_1=1/ \lambda, x_2=-1/(3\lambda)$ and plug this into the KKT condition yields $$ \lambda \left( 3\cdot \frac{1}{9\lambda^2}-3 \right)= 0 $$ Because $\lambda$ must be positive, we end up with $\lambda=+1/3$ from which you will deduce that $(3,-1)$ is the solution you are searching.

Maximizing linear objective subject to quadratic equality constraint

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