Hello, I have came up with what I think is a unique optimization problem. We are given the positive real parameters $k,t$. $t$ is the height of the rectangle, and $k$ is half of its width (see the picture). Let the parabola intersect the $x$-axis at $x=p, -p$ and pass through the top vertices: $(k,t), (-k,t)$ of the rectangle. The equation of the parabola is: $$y=ax^2+c$$ $c > t > 0$, $a < 0$. We want to find $a,c$ such that the length of the parabolic curve from $-p$ to $p$ is minimized (note that $p$ is dependent on $a$ and $c$).
My work on the problem so far: notice that $(k,t)$ is a point on the parabola. So: $$ak^2+c=t$$ $$c=t-ak^2$$ So know we really just need to find $a$. let us solve for $p$ ($x$ intersection): $$ap^2+c=0$$ $$p=\sqrt{\frac{-c}{a}}=\sqrt{\frac{ak^2-t}{a}}=\sqrt{\frac{t-ak^2}{-a}}=\frac{\sqrt{t-ak^2}}{\sqrt{-a}}$$ recall that the length of a curve from $-p$ to $p$ is given by: $$\int _{-p}^p\:\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx$$ By symmetry, the length of the parabola from $-p$ to $p$ will be 2*length of the parabola from $0$ to $p$. Subsituting: $$2\int _{x=0}^{x=p}\:\sqrt{1+\left(2ax\right)^2}dx$$ This integral is quite easy to solve with trig substitution. After solving I got that: $$f\left(a\right)=\frac{2ap\sqrt{1+4a^2p^2}+ln\left(2ap+\sqrt{1+4a^2p^2}\right)}{2a}$$ Note that $p$ is not a constant, because it is dependant on $a$. It seems as though we have made progress, we have managed to build a function, dependant on only one variable, that returns the length of the curve. The function works and returns a correct result (I have verefied it with a couple of examples) but the derivative is very long and i doubt that it is solvable. I am quite stuck here with my current knowledge. But I am positive because when you graph the function on desmos there is indeed one local minimum for $a < 0$.
An easier variant is trying to minimize the area blocked by the parabola and the $x$ axis. It was much easier to solve...