If we consider the first case, we have then
$$ -4(4-2\beta_0)+\lambda=0, $$and
$$ -2(2-\beta_1)+\lambda=0. $$which yields
$$ \beta_0=\frac{16+\lambda}{8}, $$and
$$ \beta_1=\frac{4+\lambda}{2}. $$Using the constraint on \( \beta_0 \) and \( \beta_1 \) we can then find the optimal value of \( \lambda \) for the different cases. We leave this as an exercise to you.