# Covering area as much as possible over $[0,1]^2$ keeping the average of each axis above a certain level

Consider two RV's $$X_1$$ and $$X_2$$, independently and uniformly distributed over $$[0,1]$$.

I want to find the largest possible area $$A\subset [0,1]^2$$ while keeping the average of $$X_1$$ and $$X_2$$ inside of $$A$$ above a certain level, say $$y\in(1/2,1)$$. For $$y\leq 1/2$$, it is easy to see that $$A=[0,1]^2$$ solves the problem, and that's why I excluded $$y$$'s below $$1/2$$.

So, the programming problem should be

$$\max_{A\in[0,1]^2}\int_Adx_1dx_2$$ $$s.t. E[X_1|(X_1,X_2)\in A]\geq y~\textrm{and}\\E[X_2|(X_1,X_2)\in A]\geq y.$$

Is there a general solution to this problem or any reference I can have a look to find similar types of problems?

Adding the two constraints gives $$E[X_1 + X_2 \mid (X_1, X_2) ∈ A] ≥ 2y$$. The optimal $$A$$ under only this weaker constraint is clearly the region $$x_1 + x_2 ≥ c$$ for appropriately chosen $$c$$. Specifically,
• if $$\frac12 ≤ y ≤ \frac23$$, then $$c$$ is the root of $$3 - c^3 = 3(2 - c^2)y$$ with $$0 ≤ c ≤ 1$$;
• if $$\frac23 ≤ y ≤ 1$$, then $$c = 3y - 1$$.
But since this $$A$$ is symmetric in $$x_1$$ and $$x_2$$, it happens to satisfy the original stronger constraints—so it’s optimal under them as well.
• In a parallel way, if we may consider another weaker constraint, say $E[\alpha X_1+(1-\alpha)X_2|(X_1,X_2)\in A]\geq y$, $\alpha\in(0,1)$. If this is the case, can we say $\alpha x_1+(1-\alpha)x_2\geq c$ for an appropriately chosen $c$ can also be a solution to the problem? Sep 20, 2022 at 5:41
• @Andeanlll No: if $a ≠ \frac12$, that solution won’t be symmetric in $x_1$ and $x_2$, so it will fail to satisfy the original stronger constraints. Sep 20, 2022 at 6:19