We cut the 120-cm wire into 3 pieces and from these pieces form three square frames. Can we maximize the sum of areas of these frames?

  • $\begingroup$ I think that it isn't possible because we'd have to consider a function $f(x,y)=x^2+y^2+(120-x-y)^2$ on an open square $(0, 120)^2$. $\endgroup$ – mrnobody Nov 12 '15 at 19:04
  • $\begingroup$ The areas should be $(x/4)^2$ and so on. But why do you think this is not possible? $\endgroup$ – Aretino Nov 12 '15 at 19:09
  • $\begingroup$ Thanks, my mistake :/ because the gradient of this function is $0$ just in $(40,40)$ and there is a minimum.... $\endgroup$ – mrnobody Nov 12 '15 at 19:16

Let $x$, $y$, $z$ be the edge lengths of the three frames. Then $$x\geq0, \quad y\geq 0,\quad z\geq0,\quad x+y+z=30\ .$$ This is saying that the feasible points $(x,y,z)$ make up a certain equilateral triangle $T$ in the first octant of $(x,y,z)$-space. Now you want to maximize $$q(x,y,z):=x^2+y^2+z^2$$ on $T$, which is the same thing as maximizing $\sqrt{x^2+y^2+z^2}$ on $T$. This boils down to the question: Which points of $T$ are farthest away from the origin?


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