Quick and cute summer Sangaku #3

Let the height of the rectangle be $h$ and the vertices on the blue and red parabolas be $P(x_1, h)$ and $Q(x_2,h)$ respectively. Then $h = x_1^2 \implies x_1 = \sqrt h$ and $h = \sqrt{6-x_2} \implies x_2 = 6-h^2$ . And the area of the rectangle is given $h(x_2 - x_1) = h(6-h^2-\sqrt h)$ and maximum area $A \approx 4.157551794$ $\implies \lfloor 10^7A \rfloor = \boxed{41575517}$ .

• There must be a point $\boxed{M(x;y)}$ on the blue curve and a point $\boxed{N(x';y)}$ on the red curve to construct the rectangle.
• Since point $\boxed{M(x;y)}$ is on the blue curve, we can express $\boxed{x}$ in terms of $\boxed{y}$ using the equation $\boxed{y=x^{2}}$ <=> $\boxed{x=\sqrt{y}}$
• Since point $\boxed{N(x';y)}$ is on the red curve, we can express $\boxed{x}$ in terms of $\boxed{y}$ using the equation $\boxed{y=\sqrt{6-x}}$ <=> $\boxed{x'=6-y^{2}}$
• Then the width of the rectangle is $\boxed{y}$
• The length of the rectangle is $\boxed{6-y^{2}-\sqrt{y}}$
• We can now express the rectangle's area in terms of $\boxed{y}$
• $\boxed{A(y)=y(6-y^{2}-\sqrt{y})}$
• The maximum of the function is $\boxed{4.15755179...}$
• Finally, $\boxed{\lfloor{10^{7}A}\rfloor=41575517}$