Max Area Bounded by $y^2=4ax$, $y=ax$ and $y=\frac x a$ I have the following problem which I got from an maths Instagram account that I want to solve:
https://www.instagram.com/p/CTMxl-gpfXo/?utm_source=ig_web_copy_link

Find the maximum area bounded by the curves $y^2=4ax$, $y=ax$ and $y=\frac x a$ such that $a\in [1,2]$

I have plotted the functions into the graphing calculator desmos and I have the following graph where a=1.5.

My question is what area are they referring to? The area bounded by three curves seems somewhat odd question considering you have two linear functions. Ignoring the purple function $y=\frac x a$
$$A= \int_0^{k}\sqrt{4ax}-ax dx$$
$k$ is the intersection of $y=ax$ and $y^2=4ax$ Therefore $x(a^2 x-4a)=0\Rightarrow k=4/a$
$$\Rightarrow A= \int_0^{4/a}\sqrt{4ax}-ax dx= \Big[\frac 4 3 \sqrt{(ax)^3} -\frac a 2 x^2 \Big]^{4/a}_0= \frac {32} {3a}- \frac 8 a =\frac{8}{3a} $$
It seems like a really strange problem but I want to clarify if I have done it correctly. Thank you for your time. Perhaps I should stay away from Instagram if I want to learn maths
 A: Interestingly in mathematics, straight line is also a curve (wiki)
So coming to the question, it is the area that are bound by $3$ curves, two of which are straight lines. So it should be the area that is shaded in the below diagram.

Intuitively, the bound area should increase as the difference in slopes of both straight lines increases - one of them has slope $a$ and the other has slope $\displaystyle\frac{1}{a}$.
The difference in slope is max at $a = 2$. This is also given that the aperture of parabola $y^2 = 4ax$ increases as $a$ increases.
You found the area bound between $y^2 = 4ax$ and $y = ax$, which is $ \displaystyle \frac{8}{3a}$. This area $ \displaystyle \frac{8}{3a}$ reduces as $a$ increases for $a \geq 1$.
Similarly the area bound between $y^2 = 4ax$ and $x = ay$ is,
$\displaystyle \int_0^{4a^2} \int_{y^2/4a}^{ay} dx ~ dy = \frac{8a^5}{3} ~$. This area will increase as $a$ increases (given $a \geq 1$).
So the area bound by all three curves is max at $a = 2$ and is given by -
$\displaystyle \frac{8}{3} \left(a^5 - \frac{1}{a} \right)$.
A: The region bounded by the three lines is the region above the parabola and below both straight lines. The parabola and the purple straight line meet at $\left(4a^3,4a^2\right)$ (besides $(0,0)$, of course) and the parabola and the green straight line meet at $\left(\frac4a,4\right)$ (again, besides $(0,0)$). Since $a\geqslant1$, $\frac4a\leqslant4a^3$. So, the area is equal to$$\int_0^{4/a}ax-\frac xa\,\mathrm dx+\int_{4/a}^{4a^3}\sqrt{4ax}-\frac xa\,\mathrm dx.$$
