For positive real numbers $a$ and $b$, consider the parabola $f(x)=a-bx^2$. For a value $c\in (0, \sqrt{a/b})$, we can take the tangent line to $f(x)$ at $x=c$ that encloses a triangle in the first quadrant. See the picture below, where $f(x)=14-x^2$, and we take the tangent line at $x=2$. The triangle is enclosed in purple.
Using calculus, one can show that the area of the triangle is minimized at $x=\sqrt{\frac{a}{3b}}$. It is also easy to see that $ax-bx^3$ is maximized at $x=\sqrt{\frac{a}{3b}}$. This can be seen as representing the area of rectangle in the first quadrant corresponding to the point $(x,a-bx^2)$, as pictured above in green.
While we can use calculus to show that these two optimization problems have the same answer, is there a nice explanation as to why they coincide?