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

Correct Answer: 128

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.