The maximum value of the function f(x) = x^2(60-x) in [20, 80] is:

The correct answer is Option (2) → 32000

On a closed interval, the absolute maximum is the largest of the values at the critical points and at the end points.

    \(f(x) = 60x^2 - x^3\)
    \(f'(x) = 120x - 3x^2 = 3x(40 - x)\)

\(f'(x) = 0\) gives x = 0 or x = 40; only x = 40 lies in [20, 80].

    \(f(20) = 400 \times 40 = 16000\)
    \(f(40) = 1600 \times 20 = 32000\)
    \(f(80) = 6400 \times (-20) = -128000\)

The largest of these is 32000, at x = 40.