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I am currently trying to solve a maximization problem given by

$\max_{f(x)} \int_0^1 \int_\mathbb{R} (c-y\cdot f(x)-d\cdot (x+f(x)-b)^2) \ h(x) \ dx \ dy$.

Or in other words, I have a utility function $U(f(x)):=c-y\cdot f(x)-b\cdot (x+f(x)-b)^2$ whereas $x$ and $y$ are random variables and the other deterministic variables are given. $x$ has a normal distribution (with density $h$) and $y$ is uniformly distributed on $[0,1]$. The expected utility should be the integral I'm trying to maximize above.

I have no idea how to find the optimal $f^*(x)$, since I have only worked with optimizations over variables and not functions so far. Are there any theorems or "keywords" that you could point me to that would help me solve this problem?

I once read something about a maximand being separable in a variable $x$, so choosing an optimal function $f^*(x)$ becomes equivalent to choosing an optimal $f^*$ but I don't know why that is so I am not sure if that applies to the problem above, too.

I'd be very grateful for any help! (and I apologize in advance in case I have mis-tagged this post - like I said, I am looking for the right "keywords" that could help me solve my problem)

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    $\begingroup$ The keyword is "calculus of variations". $\endgroup$ Jun 10, 2015 at 18:32
  • $\begingroup$ If you're maximizing over any function $f$, you can just solve the problem separately for each $x$, and it reduces to the standard problem where you only have an integral over $y$. Calculus of variations becomes necessary if you have some constraint on $f$ etc. $\endgroup$ Jun 10, 2015 at 19:01
  • $\begingroup$ @HenriquedeOliveira Thank you for your answer! So if I switched the order of integration (first over $x$, then over $y$) and calculated the inner integral over $y$ (which is uniformly distributed on $[0,1]$) then I can set the derivative in $f$ of the inner integral to zero and my solution would be my maximum, right? $\endgroup$ Jun 12, 2015 at 14:11
  • $\begingroup$ @DummieVariable Yes, if you have the conditions to perform those operations, that's one way to go. $\endgroup$ Jun 12, 2015 at 19:01

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