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I had a math exam today and I couldn't solve this problem. How are you supposed to calculate the maximum area of the circle c with only algebra?

urw7RSeeh8FR
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  • Can you calculate with only algebra the maximal distance between points of $;f(x)=x^2,,,g(x)=x;,;;x\in[0,1];$ ? – DonAntonio Apr 28 '17 at 12:48

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Consider the following inequality: \begin{align*} \left(x-\frac{1}{2}\right)^2\geq0\\ x^2-x+\frac{1}{4}\geq 0\\ x^2\geq x-\frac{1}{4} \end{align*} Note that equality is only reached when $x=\frac{1}{2}$. Geometrically, we can interpret this inequality as follows: the line $y=x-\frac{1}{4}$ is tangent to the graph $y=x^2$ at $x=\frac{1}{2}$.

enter image description here

After drawing a graph like the one above, it becomes apparent that the diameter of circle $C$ is at most the perpendicular distance between the two lines $y=x$ and $y=x-\frac{1}{4}$. This distance is equal to $\frac{\sqrt{2}}{8}$, making the maximum area of the circle $\frac{\pi}{128}$.

EDIT: It may be possible that the circle I mentioned before in my answer does not lie entirely above the parabola. To check this, we need to make sure that the radius of the osculating circle of the parabola at $x=\frac{1}{2}$ is greater than or equal to $\frac{\sqrt{2}}{16}$. We can calculate the radius of the osculating circle using the following formula: $$R=\frac{\left(1+y'^2\right)^{3/2}}{y''}=\frac{\left(1+2x\right)^{3/2}}{2}=\frac{2^{3/2}}{2}=\sqrt{2}>\frac{\sqrt{2}}{16}$$ This is actually not sufficient enough either, since the osculating circle at $x=\frac{1}{2}$ actually goes below the parabola:

enter image description here

Hrhm
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  • How couldn't I thought of that! Thank you so much! – urw7RSeeh8FR Apr 28 '17 at 12:52
  • @user441343 No problem. I just realized, however, that my answer is an incomplete one. It's possible that the circle I was describing in my answer is not entirely contained in the parabola (sorry for the ambiguous wording). I don't see how you could check this without calculus. – Hrhm Apr 28 '17 at 12:55
  • Problem is, you may have to show that the circle your solution exhibits doesn't cross the parabola, meaning the radius of curvature of the parabola isn't lower than the radius of the circle... – Nicolas FRANCOIS Apr 28 '17 at 12:55
  • @NicolasFRANCOIS Yes, thank you for pointing that out. I should probably edit my answer. – Hrhm Apr 28 '17 at 12:56
  • Thanks I should use that against my math teacher. – urw7RSeeh8FR Apr 28 '17 at 12:56
  • Maybe you can write the equation of the "solution" circle, find its intersections with the parabola and prove there is only one solution... – Nicolas FRANCOIS Apr 28 '17 at 12:57
  • @NicolasFRANCOIS Thanks for the proof. – urw7RSeeh8FR Apr 28 '17 at 12:59
  • But you would have to find the contact point of the tangent with the parabola, which implies some calculus, or some knowledge about how to construct a tangent to a parabola with only geometry. Not a trivial exercise... – Nicolas FRANCOIS Apr 28 '17 at 13:02