To determine the year in which the average cost of a gallon of milk [tex]\( P(t) \)[/tex] reached its maximum between 1995 and 2018, we need to find the maximum value of the function [tex]\( P(t) = -0.0004t^3 + 0.0114t^2 - 0.0150t + 2.6602 \)[/tex] in this time period.
1. Identify the Time Interval and Continuity: The interval in question is from [tex]\( t = 0 \)[/tex] (representing 1995) to [tex]\( t = 23 \)[/tex] (representing 2018). The polynomial function [tex]\( P(t) \)[/tex] is continuous and differentiable over this interval.
2. Find the Critical Points:
- To find the critical points, we first need to find the first derivative of [tex]\( P(t) \)[/tex]: [tex]\[ P'(t) = \frac{d}{dt}[-0.0004t^3 + 0.0114t^2 - 0.0150t + 2.6602] \][/tex]
- Differentiating, we get:
[tex]\[ P'(t) = -0.0012t^2 + 0.0228t - 0.015. \][/tex]
- Set the first derivative to zero to find the critical points:
[tex]\[ -0.0012t^2 + 0.0228t - 0.015 = 0. \][/tex]
- Solving this quadratic equation, we find the critical points [tex]\( t \)[/tex] in the interval [tex]\([0, 23]\)[/tex].
3. Evaluate the Second Derivative or Use the Candidates Test:
- Calculate the value of the function [tex]\( P(t) \)[/tex] at these critical points, as well as at the endpoints of the interval (1995 and 2018).
4. Evaluation and Maximization:
- Upon evaluating [tex]\( P(t) \)[/tex] at the critical points and edges of the interval, we find that [tex]\( P(t) \)[/tex] reaches a maximum value in the year 2013.
From the results of these evaluations, the year in which [tex]\( P(t) \)[/tex] reached its maximum value is [tex]\( 2013 \)[/tex]. Therefore, the answer is (2) 2013.
