maximize a function over a set using Lagrange multipliers

Question

I just learned about lagrange multipliers and I would like a feedback about my way of solving the next question, also I would like some guidance of how to continue.

Let $f: R \ ^ 3 \times R \ ^ 3 \to R$ defined by $f(u,v) = uv$ (that is $f(u,v) = \sum_{i=1}^{3}u_iv_i$ )

I need to maximize $f$ over the set $\{(u,v) \in R \ ^ 3 \times R \ ^ 3 : ||u|| \ ^ 2 =||v|| \ ^ 2 =1 \}$.

So, I defined the lagrangian:

$L(u,v,\lambda_1, \lambda_2) = f(u,v) + \lambda_1( \sum_{i=1}^{3}u_i \ ^2 -1) + \lambda_2( \sum_{i=1}^{3}v_i \ ^2 -1)$

after differentiating $f $ with respect to $u_i / v_i / \lambda_i$ I got the equations :

(1) $v_i +2u_i\lambda_1 = 0$

(2) $u_i +2v_i\lambda_2 = 0$

(3)$\sum_{i=1}^{3}u_i \ ^2 =1$

(4)$\sum_{i=1}^{3}v_i \ ^2 =1$

Now, $\lambda_i \ne 0$ because $u \ne 0 ,v \ne 0$.

Im not sure how to continue, and im supposed to get that the max/min is at $u=v , u=-v$ which im not sure im getting here.

Thanks for helping!

Answer 1

Hint: You've set up the equations correctly, so what remains is to solve the system. From your first equation, you have $v_i=-2u_i\lambda_1$. If you substitute this into the fourth equation and use the third to simplify, you can solve for $\lambda_1$. A similar route works for $\lambda_2$.