Question

Let $A$ be the region enclosed by the parabola $y^2 = 2x$ and the line $x = 24$. Then the maximum area of the rectangle inscribed in the region $A$ is ________.

Answer 1

Consider a rectangle inscribed in the region bounded by the parabola $y^2 = 2x$ and the line $x = 24$. Let the coordinates of the upper right corner of the rectangle be $\left(\frac{b^2}{2}, b\right)$, where $b$ is the $y$-coordinate of the corner on the parabola.
The length of the rectangle along the $x$-axis is:
$2 \left(24 - \frac{b^2}{2}\right)$.
The height of the rectangle is:
$b$.
Therefore, the area $A$ of the rectangle is given by:
$A = 2 \left(24 - \frac{b^2}{2}\right) \times b$.
Simplifying:
$A = 2 \left(24b - \frac{b^3}{2}\right)$,
$A = 48b - b^3$.
To find the maximum area, we differentiate $A$ with respect to $b$ and set the derivative equal to zero:
$\frac{dA}{db} = 48 - 3b^2 = 0$.
Solving for $b$:
$3b^2 = 48$,
$b^2 = 16$,
$b = 4$ (since $b>0$).
Substituting $b = 4$ back into the expression for $A$:
$A = 2 \left(24 - \frac{4^2}{2}\right) \times 4$,
$A = 2 \times (24 - 8) \times 4$,
$A = 2 \times 16 \times 4$,
$A = 128$.
Therefore, the maximum area of the rectangle is:
128.