Find the global maximum and minimum values of the function on the given interval. State as ordered pairs.
$y=8+27x-x^3;[0,4]$

• A. Global maximum at $\(\left\)(3,62\(\right\))$, Global minimum at $\(\left\)(0,8\(\right\))$
• B. Global maximum at $\(\left\)(3,62\(\right\))$, Global minimum at $\(\left\)(4,52\(\right\))$
• C. Global maximum at $\(\left\)(-3,-46\(\right\))$, Global minimum at $\(\left\)(0,8\(\right\))$
• D. Global maximum at $\(\left\)(4,52\(\right\))$, Global minimum at $\(\left\)(-3,-46\(\right\))$

Answer: (A)

Verified step by step guidance

1. Step 1: Understand the problem. We are tasked with finding the global maximum and minimum values of the function y = 8 + 27x - x^3 on the interval [0, 4]. This involves evaluating the function at critical points and endpoints within the interval.
2. Step 2: Find the derivative of the function to locate critical points. The derivative of y = 8 + 27x - x^3 is computed as dy/dx = 27 - 3x^2. Set this derivative equal to zero to find critical points: 27 - 3x^2 = 0.
3. Step 3: Solve for x in the equation 27 - 3x^2 = 0. Rearrange to get x^2 = 9, then take the square root to find x = 3 and x = -3. Since the interval is [0, 4], only x = 3 is within the interval.
4. Step 4: Evaluate the function y = 8 + 27x - x^3 at the critical point x = 3 and at the endpoints x = 0 and x = 4. This will give the function values at these points, which are candidates for the global maximum and minimum.
5. Step 5: Compare the function values obtained in Step 4. The largest value corresponds to the global maximum, and the smallest value corresponds to the global minimum. State the results as ordered pairs (x, y).