Area of square under a curve.

Question

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.

Answer 1

There have to be reasons if you are "on the verge of getting banned". E.g., we are not given any hint as to the source of this quite involved problem. Anyway, this problem is of a purely algebraic nature. Please don't write "$P(A^{1/3},0)$" when you mean "$P=(A^{1/3},0)$". No function $P$ of two variables is defined in the actual context.

The problem obviously has exactly one solution $(a,b)$. It is therefore sufficient to argue about necessary conditions for $a$ and $b$.

The quantities $a\ne b$ satisfy $$b-a=a(3-a^2),\qquad b-a=b(3-b^2)\ .\tag{1}$$ Subtracting these equations we obtain $0=(b-a)(3-a^2-ab-b^2)$, hence $$3-a^2-b^2=ab\>,\quad{\rm resp.,}\quad (a+b)^2=3+ab\ .\tag{2}$$ Adding the equations $(1)$ leads together with $(2_1)$ to $$2(b-a)=(a+b)(3-a^2+ab-b^2)=2(a+b)ab,$$ hence $b-a=(a+b)ab$. Squaring this we obtain $(b-a)^2=(a+b)^2(ab)^2$, hence $$(a+b)^2-4ab=(a+b)^2(ab)^2\ ,\tag{3}$$ and together with $(2_2)$: $$3(1-ab)=(3+ab)(ab)^2\ .$$ Now this is saying that $(ab+1)^3=4$, hence $z:=ab=2^{2/3}-1$. Now $(3)$ gives $$(a+b)^2={4z\over1-z^2}$$ and therefore $$(b-a)^2=(a+b)^2-4ab={4z^3\over1-z^2}=6-3\cdot2^{2/3}=\root3 \of{216}-\root3\of {108}\ .$$ It follows that $A=216$ and $B=-108$. The rest is easy.

Answer 2

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.

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: $$\bbox[5px,border:2px solid black]{f(q)^2 \iff f\left(\sqrt{3-\sqrt[3]{2}-2^{2/3}}\right)^2=6-3\times 2^{2/3}}$$

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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: $$\therefore{A=216\\B=-108}$$

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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 $$r=\sqrt{3^2+\left(-\frac{3\times2^{2/3}}{2}\right)^2}=\sqrt{9+\frac{9}{2^{2/3}}}=\frac{3 \sqrt{1+2^{2/3}}}{\sqrt[3]{2}}$$