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

The correct answer is Option (2) → 32000

Given: $f(x) = x^2 (60 - x) = 60x^2 - x^3$

Derivative: $f'(x) = 120x - 3x^2 = 3x(40 - x)$

Set $f'(x) = 0 \Rightarrow x = 0 \text{ or } x = 40$

Check interval $[20,80]$: critical point $x = 40$ and endpoints $x = 20, 80$

Compute $f(x)$:

$f(20) = 20^2 (60-20) = 400*40 = 16000$

$f(40) = 40^2 (60-40) = 1600*20 = 32000$

$f(80) = 80^2 (60-80) = 6400*(-20) = -128000$

Maximum value: $f_{\max} = 32000$ at $x=40$