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Let $D\subseteq [0,2\pi]\times [0,2\pi]$ and ${D}^\complement$ be the complementary region, i.e.

  1. $D \cup {D}^\complement = [0,2\pi]\times [0,2\pi]$ and
  2. $D\cap {D}^\complement = \emptyset$.

I would like to find $D$ such that I maximize the quantity $$ \int_D \cos(x)\mathrm{d}x \mathrm{d}y-\int_{{D}^\complement} \cos(x)\mathrm{d}x \mathrm{d}y$$ while $$\int_D \cos(y)\mathrm{d}x \mathrm{d}y-\int_{{D}^\complement} \cos(y)\mathrm{d}x \mathrm{d}y>T$$ for some value $T$.

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  • $\begingroup$ Just a small contribution, but you have that $\int_{D^c}\cos(x)\mathrm{d}x \mathrm{d}y=\int_0^{2\pi}\int_0^{2\pi}\cos(x)\mathrm{d}x \mathrm{d}y-\int_D \cos(x)\mathrm{d}x \mathrm{d}y=-\int_D \cos(x)\mathrm{d}x \mathrm{d}y$, so the problem becomes maximizing $2\int_D \cos(x)\mathrm{d}x \mathrm{d}y$ subject to $2\int_D \cos(y)\mathrm{d}x \mathrm{d}y>T$. $\endgroup$
    – Varun Vejalla
    Commented Aug 2, 2022 at 20:42
  • $\begingroup$ The problem comes down to finding $g(y):[0,2\pi]\to[-\pi/2,\pi/2]$ such that $4\int_0^{2\pi}\cos(g(y))dy$ is maximized subject to the constraint $4\int_0^{2\pi}\cos(y)g(y)dy=T$. The region $D$ would then be $D=\{(x,y)\mid \frac{\pi}{2}-g(y)\le |x-\pi|\le \pi, 0\le y \le 2\pi\}$. I believe $g(y)=\arcsin(\lambda \cos(y))$, obtained using the technique here, is the maximizing function, although I'm not sure I did it correctly. $\endgroup$
    – Varun Vejalla
    Commented Aug 3, 2022 at 5:54

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