On the interval , the function takes its maximum value at the point
- A.0
- B.1/4
- C.1/2
- D.1/3
Show answer and solution
Correct option: B
Solution
Differentiate f(x) = x25·(1-x)75 and set to zero:
f'(x) = 25x24·(1-x)75 - 75x25·(1-x)74 = x24·(1-x)74·[25(1-x) - 75x] = 0
Setting the bracket to zero: 25 - 25x - 75x = 0 → 25 = 100x → x = 1/4.
This is the only interior critical point, and since f(0) = f(1) = 0 while f(1/4) > 0, the maximum occurs at x = 1/4.
f'(x) = 25x24·(1-x)75 - 75x25·(1-x)74 = x24·(1-x)74·[25(1-x) - 75x] = 0
Setting the bracket to zero: 25 - 25x - 75x = 0 → 25 = 100x → x = 1/4.
This is the only interior critical point, and since f(0) = f(1) = 0 while f(1/4) > 0, the maximum occurs at x = 1/4.
[1995]