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

The function $f(x) = x^2(60 - x)$ needs to be evaluated on the interval [20, 80]. The maximum can occur at critical points inside the interval or at the endpoints.

---

Expanding the function:

$f(x) = x^2(60 - x)$

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

Taking the derivative:

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

$f'(x) = 3x(40 - x)$

Setting the derivative equal to zero:

$3x(40 - x) = 0$

This gives $x = 0$ or $x = 40$.

Since the interval is [20, 80], only $x = 40$ is relevant.

---

Evaluating at $x = 20$:

$f(20) = (20)^2(60 - 20)$

$f(20) = 400 \times 40$

$f(20) = 16000$

---

Evaluating at $x = 40$:

$f(40) = (40)^2(60 - 40)$

$f(40) = 1600 \times 20$

$f(40) = 32000$

---

Evaluating at $x = 80$:

$f(80) = (80)^2(60 - 80)$

$f(80) = 6400 \times (-20)$

$f(80) = -128000$

---

Comparing the values: $f(20) = 16000$, $f(40) = 32000$, and $f(80) = -128000$.

Therefore, the maximum value of the function on [20, 80] is $32000$.