# Maximization of 4-variable function...

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?

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.
$$y_1-x_1 \leq \sqrt {y_1^{2}+x_1^{2}} \sqrt {1+1}=\sqrt 2$$. 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 $$y_2=-\frac 1 {\sqrt 2}$$.
• Thank you for the answer, but could you please explain where the $\sqrt{2}$ comes from? Also, what if we placed the restriction that $x_1,x_2,y_1,y_2 > 0$? Sep 20 '19 at 23:57
• @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 $x_2=1,x_1=0, y_1=1,y_2=0$/. Sep 21 '19 at 0:03