Given
$\min x_1^2+2x_2^2+x_1$
s.t : $x_1+x_2\leq a$
Prove:
a)for every $a\in\mathbb{R}$ the problem has a unique solution.
b)find the optimal solution as a function of $a$.
c)let f(a)be optimal value find explict formula for f(a) and prove it is convex.
So for a) i said the the target function is strictly convex and the constrain is convex hence we have a unique optimal solution.
for b)I found that $x_1=\frac{4a-1}{6},x_2=\frac{2a+1}{6}$ (not sure if it's right)
for c) I didn't understand what i'm suppose to do.
Hints/answers will be welcome.
The function $f(a)$ is given by $$ f(a) = \max_{x_1,x_2} x_1^2 + 2x_2^2 + x_1 + \mu(x_1+x_2-a). $$ The Envelope Theorem is that you can differentiate the Lagrangian wrt the parameter of interest, then evaluate at the solution to see how the optimized solution varies with the parameter (https://en.wikipedia.org/wiki/Envelope_theorem). So $$ f'(a) = -\mu^* $$ Now, to prove convexity, you have to show that $f'(a)$ is increasing in $a$. That is equivalent to showing that the multiplier on the constraint, $\mu^*$, is increasing in $a$. If you increase $a$, you expand the choice set, and the objective must decreasing, so $\mu^*$ is decreasing in $a$. That implies $f''(a) = -D_a \mu^*(a) \ge 0$, and $f$ is convex.
Or you could plug your optimal solution back into the objective, and differentiate wrt $a$ and see what you get.