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Maximize $$x_2 - x_1 + y_1 - y_2$$ given that $x_1^2 + y_1^2 =1$ and $x_2^2 + y_2^2 = 1$.

I was thinking about using Lagrange multipliers, but I only know how that works for a 3-variable function, not 4. Could someone please suggest a way to solve this? Maybe with Lagrange multipliers or some more elementary method?

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By the hypotesis you can write $x_1=\sin \theta, y_1=\cos \theta$ and $x_2=\sin \alpha, y_2=\cos \alpha$. Then, your want to find the maximum value of $$E=(\sin \alpha - \sin \theta)+(\cos \alpha - \cos \theta)=(\sin \alpha+\cos \alpha) -(\sin \theta +\cos \theta).$$ But, $-\sqrt{2}\le \sin x+\cos x\le \sqrt{2}, \ \forall x\in [0,2\pi]$ and the equality holds for $\alpha=\pi/4$ and $\theta=5\pi/4$. In particular, $E\le 2\sqrt{2},$ exactly as professor Rama Murty found.

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$y_1-x_1 \leq \sqrt {y_1^{2}+x_1^{2}} \sqrt {1+1}=\sqrt 2$. Similarly $x_2 -y_2\leq \sqrt 2$ so the given expession does not exceed $2\sqrt 2$. To see that this value is actually attained take $x_1=-\frac 1 {\sqrt 2}$, $y_1=\frac 1 {\sqrt 2}$ $x_2=\frac 1 {\sqrt 2}$ and $y_2=-\frac 1 {\sqrt 2}$.

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  • $\begingroup$ Thank you for the answer, but could you please explain where the $\sqrt{2}$ comes from? Also, what if we placed the restriction that $x_1,x_2,y_1,y_2 > 0$? $\endgroup$ Sep 20 '19 at 23:57
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    $\begingroup$ @OmicronGamma I have used Cauchy- Schwarz inequality $|ab+cd| \leq (a^{2}+b^{2})^{1/2} (c^{2}+d^{2})^{1/2}$ with $c=d=1$. If you restrict $x_i,y_i, i=1.2$ to psoitive values then the maximum is $2$ and it is attained when $x_2=1,x_1=0, y_1=1,y_2=0$/. $\endgroup$ Sep 21 '19 at 0:03

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