Question Detail
Question
The function y = 9x³ – 45x² + 48x + 11 has a maximum value at y = ?
Correct Answer
Option C is the correct answer.
Detailed Explanation
- Find the first derivative: dy/dx = 27x² – 90x + 48.2. Set dy/dx = 0 to find critical points: 27x² – 90x + 48 = 0. Divide by 3: 9x² – 30x + 16 = 0.3. Solve for x using the quadratic formula: x = [30 ± √(900 - 4916)] / (2*9) = [30 ± √(900 - 576)] / 18 = [30 ± √324] / 18 = [30 ± 18] / 18. Critical points are x₁ = (30 + 18)/18 = 48/18 = 8/3 and x₂ = (30 - 18)/18 = 12/18 = 2/3.4. Find the second derivative: d²y/dx² = 54x – 90.5. Apply the second derivative test: - At x = 8/3: d²y/dx² = 54(8/3) - 90 = 144 - 90 = 54 > 0 (Local Minimum). - At x = 2/3: d²y/dx² = 54(2/3) - 90 = 36 - 90 = -54 < 0 (Local Maximum).6. Substitute x = 2/3 into the original function to find the maximum y-value: y = 9(2/3)³ – 45(2/3)² + 48(2/3) + 11 y = 9(8/27) – 45(4/9) + 32 + 11 y = 8/3 – 20 + 43 = 8/3 + 23 = 8/3 + 69/3 = 77/3.This matches option C.
Hint
To find extrema, first find the critical points by setting the first derivative to zero. Then use the second derivative test to distinguish between maximum and minimum. Finally, plug the x-value back into the original function.