[Math] Area of square under a curve.

analytic geometry

A square having sides parallel to the coordinate axes is inscribed in the region.
{$(x,y):x,y>0:y\le -x^3+3x$}.
If the area of the square is written as $A^{1/3}+B^{1/3}$ square units where $A,B\in \Bbb Z$ and $A>B$, then find
(i)$\sqrt{A-B}$
(ii)Slope of line with x and y intercepts as $A,B$ respectively
(iii)$A+B\over A-B$
(iv)Circumradius of $\triangle OPQ$ where $O$ is origin, $P(A^{1/3},0)$ and $Q(0,B^{1/3})$

My approach:
Let the square be $KLMN$ where $K,L$ lie on x-axis with coordinates $(a,0);(b,0)$ respectively.
This will give coordinates of other points i.e. $M,N$ in terms of $a,b$.
Now taking out the length of all sides and equating them will give $a,b$.

Problem with my approach:
I end up in a cumbersome equation involving $a,b$

Tips: I know nothing more than simple differentiation and limits. So please avoid solutions with integration and other stuffs.

PS: No need to downvote as I'm on the verge of getting banned. Please comment if you have any problem with the question.

enter image description here

Best Answer

Define $f(x):=3 x - x^3$. Obviously, the square must have a side length of $f(q)$, and we need to find $q$ that gives the largest square inside $f(x)$.

The line $y=f(q)$ intersects $y=f(x)$ at three points: $$\left(q,f\left(q\right)\right)\\ \left(\frac{1}{2} \left(-\sqrt{3} \sqrt{4-q^2}-q\right),f\left(\frac{1}{2} \left(-\sqrt{3} \sqrt{4-q^2}-q\right)\right)\right)\\ \color{red}{\left(\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right),f\left(\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right)\right)\right)}\\ $$

We are only interested at the last point. The top left vertex of the square is $(q,f(q))$, and we need to ensure that $\sqrt{(f(q)-f(z))^2+(q-z)^2}=f(q)$, where $z=\frac{1}{2} \left(\sqrt{3} \sqrt{4-q^2}-q\right)$. So we need to find $q$ for the equation below: $$\sqrt{(f(q)-f(z))^2+(q-z)^2}-f(q)=0\tag{1}$$

You can use this implementation to get the intuition for the solution. enter image description here

This gives us two solutions: $$\left\{\left\{q\to \sqrt{1+\sqrt[3]{2}}\right\},\left\{q\to \sqrt{3-\sqrt[3]{2}-2^{2/3}}\right\}\right\}$$ Which means the area of the maximum square is: $$\bbox[5px,border:2px solid black]{f(q)^2 \iff f\left(\sqrt{3-\sqrt[3]{2}-2^{2/3}}\right)^2=6-3\times 2^{2/3}}$$

To get $A$ and $B$ set: $$A^{1/3}=6\iff A=6^3=216\\ B^{1/3}=3\times2^{2/3} \iff B=(3\times2^{2/3})^3=108$$ Since $\sqrt[3]A+\sqrt[3]{-B}=\sqrt[3]A-\sqrt[3]{B}$, for reals, we have the solution: $$\therefore{A=216\\B=-108}$$

The questions can now be easily answered:

  1. $\sqrt{216+108}=\sqrt{324}=18$
  2. Line passes through $(216,0)$ and $(0,-108)$, therefore: $y=\frac{-108}{216}(x-216)\iff y=\frac12x+108$
  3. $\frac{108}{324}=\frac13$
  4. The circumcircle is centered at: $$\left(\frac62,-\frac{3\times2^{2/3}}{2}\right)$$ therefore the radius is $$r=\sqrt{3^2+\left(-\frac{3\times2^{2/3}}{2}\right)^2}=\sqrt{9+\frac{9}{2^{2/3}}}=\frac{3 \sqrt{1+2^{2/3}}}{\sqrt[3]{2}}$$
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