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Question:

Let $0<x<\dfrac{3}{2}$, then find the maximum of $$\sqrt{x^2+x^4}+\sqrt{\left(x-\dfrac{3}{2}\right)^2+\left(x^2-\dfrac{9}{4}\right)^2}$$

I Found when $x=\dfrac{1}{2}$ it's maximum,see the maximum .

This result seems interesting, so I think there exist simple methods.

My idea: Let $$f(x)=\sqrt{x^2+x^4}+\sqrt{\left(x-\dfrac{3}{2}\right)^2+\left(x^2-\dfrac{9}{4}\right)^2}$$ then I found $f'(x)$ is very ugly,then I can't find this roots by hand.

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  • $\begingroup$ Please change every occurrence of maximum to minimum. $\endgroup$ – Siminore Sep 6 '14 at 10:02
  • $\begingroup$ @Siminore. No. He is restricting $x$ to (0,1.5). Over that range the maximum probably is at $x=0.5$ as he says. $\endgroup$ – almagest Sep 6 '14 at 10:12
  • $\begingroup$ Oh sorry, I was looking at the whole graph in the link! $\endgroup$ – Siminore Sep 6 '14 at 10:31
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A geometric solution:

Note that the given expression equals sum of distances from $P = (x,x^2)$ to $(0,0)$ and $(\frac{3}{2},\frac{9}{4})$. These are points on the parabola $y = x^2$.

If you ignore the constraint that $P$ has to lie on parabola, the locus of points for which the sum is constant is a ellipse with foci $(0,0)$ and $(\frac{3}{2},\frac{9}{4})$. Maximising sum thus equals maximising radius of the ellipse such that the ellipse still intersects parabola. (Since we are working on closed interval, this maximum exists.)

But then ellipse has to be tangent to parabola (otherwise, essentially, radius of the ellipse could be increased a bit). Now, since tangent of ellipse reflects ray from the other focus to another (see Reflective Properties of Ellipse), tangent of parabola must also reflect ray from $(0,0)$ to $(\frac{3}{2},\frac{9}{4})$, which then again means that reflection of $(0,0)$ with respect to tangent line at $(x,x^2)$, $(x,x^2)$ itself, and $(\frac{3}{2},\frac{9}{4})$ are collinear.

Now, you can work out that the reflection point is $$ \left( \frac{4x^3}{4x^2+1}, -\frac{2x^2}{4x^2+1} \right). $$ Checking the collinearity is easy; slopes have to match. After some manipulation you are left out with $$ \frac{3}{2} = 4x^3+2x $$ for which the unique real root is $\frac{1}{2}$ which therefore give the maximum.

It requires some work, but it gives also the general solution when $\frac{3}{2}$ and $\frac{9}{4}$ are replaced by $a$ and $a^2$ (just replace $\frac{3}{2}$ with $a$ in the last equation) and an immediate answer for the search of the minimum for the original function (triangle inequality).

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