# maximum area of inscribed quadrilateral

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.

• Show us what you've done. Otherwise, how can we explain why it's right? – saulspatz May 2 at 11:21
• What kind of quadrilateral is it? Arbitrary or some special? – Aqua May 2 at 11:21
• It is arbitrary. Thank you for asking. – macrow May 2 at 11:25
• I proved with expressions that two points of the rectangle must be placed where the graph meets with x-axis. Also we can explain with specifying rectangle's point with proof and explanation. – macrow May 2 at 11:34
• You could find the area under the curve between x=0 and x=2 by integration. This area will be the upper bound of the inscribed shape of which has an area described by the formula geeksforgeeks.org/maximum-area-quadrilateral or here:mathopenref.com/quadrilateralinscribedarea.html also max are is shown here: en.wikipedia.org/wiki/… – NoChance May 2 at 11:41

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): $$S=\frac12\cdot a\cdot (2a-a^2)+\frac12\cdot (2a-a^2+2b-b^2)(b-a)+\frac12\cdot (2-b)(2b-b^2)=\\ -\frac12a^2b+\frac12ab^2-b^2+2b,0 Now maximize $$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.