maximum area of inscribed quadrilateral

Question

Find the maximum area of the quadrilateral inscribed on $y=2x-x^2$, where $y\geq 0$ and explain your answer.

I can just estimate the shape but I don't know how to prove it precisely. Help me with a explanation please.

Answer 1

For arbitrary quadrilateral, obviously, the two vertices should lie at $(0,0)$ and $(2,0)$ and the other two should lie on different sides of $(1,0)$. Refer to the figure:

Let the two vertices be $A(a,2a-a^2)$ and $B(b,2b-b^2)$. Note the quadrilateral is not necessarily an isosceles trapezium (although it will be at the end). Find the area of the quadrilateral (which is the sum of the areas of two triangles on the sides and trapezium in the middle): $$S=\frac12\cdot a\cdot (2a-a^2)+\frac12\cdot (2a-a^2+2b-b^2)(b-a)+\frac12\cdot (2-b)(2b-b^2)=\\ -\frac12a^2b+\frac12ab^2-b^2+2b,0<a<1<b<2$$ Now maximize $S$: $$\begin{cases}S_a=-ab+\frac12b^2=0\\ S_b=-\frac12a^2+ab-2b+2=0\end{cases}\Rightarrow \begin{cases}a=\frac23\\ b=\frac43\end{cases}$$ I will leave checking the Hessian for you. Hence, $S(\frac23,\frac43)=\frac{32}{27}$ is maximum area.

Note that the equations of red, green, blue lines are: $y=\frac43x,y=\frac89,y=-\frac43x+\frac83$, respectively. So, the quadrilateral is an isosceles trapezium.