(a) Set \(y = 0\) to find the x-coordinate of A:

\(9(x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}) = 0\)

\(9x^{-\frac{3}{2}}(x - 4) = 0\)

\(x = 4\) only.

(b) Differentiate \(y\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 9\left(-\frac{1}{2}x^{-\frac{3}{2}} + 6x^{-\frac{5}{2}}\right)\)

At \(x = 4\), gradient = \(9\left(-\frac{1}{16} + \frac{6}{32}\right) = \frac{9}{8}\)

Equation of tangent: \(y = \frac{9}{8}(x - 4)\)

(c) Set \(\frac{dy}{dx} = 0\) to find maximum point:

\(9x^{-\frac{5}{2}}\left(-\frac{1}{2}x + 6\right) = 0\)

\(x = 12\)

(d) Integrate to find the area:

\(\int_4^9 9(x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}) \, dx = 9\left[\frac{x^{\frac{1}{2}}}{\frac{1}{2}} - \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}\right]_4^9\)

\(= 9\left[(6 + \frac{8}{3}) - (4 + 4)\right] = 6\)