The maximum value of the function f(x) = 2x³ - 5x² + 36x on the set A = {x | x² + 20 ≤ 9} is

x² + 20 ≤ 9
x² ≤ -11
This inequality has no real solutions, since x² is always non-negative. However, there's likely a typo in the question. Let's assume the inequality was meant to be x² - 20 ≤ 9. Then:
x² - 20 ≤ 9
x² ≤ 29
-√29 ≤ x ≤ √29
Approximately -5.38 ≤ x ≤ 5.38
Let's assume the intended inequality was x² - 20 ≤ 9, which simplifies to x² ≤ 29. Then -√29 ≤ x ≤ √29.
Let's proceed assuming the correct inequality is x²-20 ≤ 9 which gives -√29 ≤ x ≤ √29.
Now, f(x) = 2x³ - 5x² + 36x
Differentiate wrt x:
f'(x) = 6x² - 10x + 36 = 2(3x² - 5x + 18)
To find critical points, we set f'(x) = 0:
3x² - 5x + 18 = 0
The discriminant is b² - 4ac = (-5)² - 4(3)(18) = 25 - 216 = -191, which is negative. Therefore, there are no real roots, meaning f'(x) is always positive (since the parabola opens upwards). This implies that f(x) is strictly increasing.
Since f(x) is strictly increasing, the maximum value will occur at the upper bound of the interval. Considering -√29 ≤ x ≤ √29, the maximum occurs at x = √29.
f(√29) = 2(√29)³ - 5(√29)² + 36(√29) ≈ 2(24.37) - 5(29) + 36(5.38) ≈ 48.74 - 145 + 193.68 ≈ 197.42
If we assume the inequality was x² + 20 ≤ 9, then x² ≤ -11, which has no real solution. The problem statement might contain errors.