The function f(x)=x2(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)=x2(60−x)
f(x)=60x2−x3
Taking the derivative:
f′(x)=120x−3x2
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)=400×40
f(20)=16000
Evaluating at x=40:
f(40)=(40)2(60−40)
f(40)=1600×20
f(40)=32000
Evaluating at x=80:
f(80)=(80)2(60−80)
f(80)=6400×(−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.