The parabola defined by the equation $y^{2}=2 p x$ is bounded by the line $x=a$. Into this parabolic segment, we inscribe rectangles whose midline is the axis of the parabola. Which rectangle will have the maximum area?
Let the coordinates of the vertices of the rectangle lying on the parabola be $x$ and $y= \pm \sqrt{2 p x}$. From the diagram, it can be read that the area of the sought rectangle is

$$
T=2(a-x) \sqrt{2 p x}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_6fb5cd5f3e48a538b646g-1.jpg?height=586&width=631&top_left_y=325&top_left_x=736)

This has its maximum where

$$
\frac{T^{2}}{4 p}=2 x(a-x)^{2}=2 x(a-x)(a-x)
$$

The sum of the three factors on the right is the constant $a$, so according to the well-known inequality between the arithmetic and geometric means, the product will be maximal when the factors are equal, that is,

$$
2 x=a-x, \quad x=\frac{a}{3}
$$

From the equation of the parabola, the corresponding ordinates are: $y= \pm \sqrt{\frac{2 a p}{3}}$, and thus the maximum area is $T_{\max }=\frac{4 a}{3} \sqrt{\frac{2 a p}{3}}$.