algebra-precalculuscalculusmaxima-minima
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?
One way is to note that $x_1y_2 - x_2y_1$ is a signed area, i.e. it may positive or negative. Adding up all the signed areas of the triangles formed by the points $O$, $P_k$ and $P_{k+1}$ will cancel all the superfluous parts, as can be seen from the sketch:
The same argument also works for trapezoids with the $x$-axis instead of triangles with the origin. (I think this is the most illuminating argument, because the key trick is to give the area a sign depending on orientation.)
One could argue that this is not very rigorous, however. A more rigorous proof is to divide the polygon into two smaller polygons (it's not trivial to show that this is possible) and argue that adding the shoelace sums of the two parts gives the shoelace sum of the whole. That's because the two additional terms for the extra side cancel each other. (This cancellation is well-known from line integrals, we are in essence calculating $\frac12 \oint_{polygon} x\,dy - y\,dx$ here.) By induction, you then only have to verify the formula for a triangle.
Too long for a comment.
You missed a term in $J^{*}$; it should be $J^{*}=\int_0^{\pi/2}F^{*}(x,y_1,y_2,y_1',y_2',\lambda)\,dx$, where $$ F^{*}(x,y_1,y_2,y_1',y_2',\lambda)=2y_1y_2+(y_1')^2+(y_2')^2+\lambda(y_1'+y_2'-4x). \tag{1} $$ Now, write the Euler-Lagrange equations for $F^{*}$: \begin{align} &\frac{\partial F^{*}}{\partial y_1}-\frac{d}{dx}\left(\frac{\partial F^{*}}{\partial y_1'}\right)=0\implies 2y_2-2y_1''-\lambda'=0, \tag{2a} \\ &\frac{\partial F^{*}}{\partial y_2}-\frac{d}{dx}\left(\frac{\partial F^{*}}{\partial y_2'}\right)=0\implies 2y_1-2y_2''-\lambda'=0, \tag{2b} \\ &\frac{\partial F^{*}}{\partial \lambda}=0\implies y_1'+y_2'-4x = 0. \tag{2c} \end{align} Can you continue from here?
Best Answer
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.