Crazy Areas!

The graph of the three curves is shown above.

Let the parabola $y^2 = 4ax$ cuts line $y=ax$ and $y= \dfrac xa$ at $(x_1, y_1)$ and $(x_2, y_2)$ respectively. Then we have:

$\begin{cases} y_1^2 = 4ax_1 & \implies (ax_1)^2 = 4ax_1 & \implies x_1 = \dfrac 4a & \implies y_1 = 4 \\ y_2^2 = 4ax_2 & \implies \left(\dfrac {x_2}a \right)^2 = 4ax_2 & \implies x_2 = 4a^3 & \implies y_2 = 4a^2 \end{cases}$

From the graph, we note that the area bounded by the three curves is given below:

$\begin{aligned} A & = \underbrace{\frac {x_1y_1}2}_{\text{triangle}} + \underbrace{\int_{x_1}^{x_2} \sqrt{4ax} \ dx}_{\text{parabola}} - \underbrace{\frac {x_2y_2}2}_{\text{triangle}} \\ & = \frac 8a + 2 \sqrt a \int_\frac 4a^{4a^3} x^\frac 12 dx - 8a^5 \\ & = \frac 8a + 2 \sqrt a \cdot \frac 23 x^\frac 32 \bigg|_\frac 4a^{4a^3} - 8a^5 \\ & = \frac 8a + \frac {4 \sqrt a} 3 \left(8a^\frac 92 - 8a^{-\frac 32} \right) - 8a^5 \\ & = \frac 8a + \frac {32}3 \left(a^5 - \frac 1a \right) - 8a^5 \\ & = \left(\frac {32}3 - 8\right) \left(a^5 - \frac 1a \right) \\ & = \frac 83 \left(a^5 - \frac 1a \right) \end{aligned}$

For $a \in [1, 2]$ , $a^5 - \dfrac 1a$ increases with $a$ . Therefore, the largest area $A_{max} = \dfrac 83 \left(2^5 - \dfrac 12 \right) = \dfrac 83 \left(32 - \dfrac 12 \right) =\boxed{84}$ .