Find the largest value of $x$ for which $x^2 + y^2 + z^2 = x + y + z$.

What I did was subtract the RHS, to get $$x^2 - x + y^2 - y + z^2 - z = 0$$

$$x^2 - x + \frac{1}{4} + y^2 - y + \frac{1}{4} + z^2 - z + \frac{1}{4} = \frac{3}{4}$$

$$(x-\frac{1}{2})^2 + (y-\frac{1}{2})^2 + (z-\frac{1}{2})^2 = \frac{3}{4}$$

Hence the given equation can be rewritten as the formula for a sphere centered at $(x,y,z) = (\frac{1}{2},\frac{1}{2},\frac{1}{2})$ with a radius of $\frac{\sqrt3}{2}$.

Now for $x$ to be at its largest value, that would mean $y = z = \frac{1}{2}$ (right?).

And if that is true, then $\boxed{x = \frac{\sqrt{3}+1}{2}}$.

Please let me know if I am correct, and if not, please help me understand how to achieve the correct answer. Thanks!

Edit: Thanks everyone for your responses!

  • 3
    $\begingroup$ Looks right to me. $\endgroup$ – user164587 Oct 11 '14 at 2:53

Yes, you are correct. The point where $(y - a)^2$ and $(z - b)^2$ are zero is where we would find the maximum $x$. Anything greater than zero would make the $x$ part of the equation smaller.

Nice thinking!


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