The first Case

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.

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