To determine the year between 1995 and 2018 when the average cost of a gallon of milk in the United States reached its maximum, we need to analyze the given function [tex]\( P(t) = -0.0004t^3 + 0.0114t^2 - 0.0150t + 2.6602 \)[/tex], where [tex]\( t \)[/tex] is the number of years since January 1995. Let's proceed step-by-step:
### Step 1: Find the derivative of [tex]\( P(t) \)[/tex]
The maximum value of [tex]\( P(t) \)[/tex] occurs at the critical points, which we find by taking the derivative of [tex]\( P(t) \)[/tex] and setting it equal to zero.
[tex]\[ P'(t) = \frac{d}{dt} \left( -0.0004t^3 + 0.0114t^2 - 0.0150t + 2.6602 \right) \][/tex]
Using basic differentiation rules:
[tex]\[ P'(t) = -0.0012t^2 + 0.0228t - 0.015 \][/tex]
### Step 2: Set the derivative equal to zero and solve for [tex]\( t \)[/tex]
[tex]\[ -0.0012t^2 + 0.0228t - 0.015 = 0 \][/tex]
This is a quadratic equation in the standard form [tex]\( at^2 + bt + c = 0 \)[/tex]. To solve for [tex]\( t \)[/tex], we use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation:
- [tex]\( a = -0.0012 \)[/tex]
- [tex]\( b = 0.0228 \)[/tex]
- [tex]\( c = -0.015 \)[/tex]
Plugging in the values, we get:
[tex]\[ t = \frac{-0.0228 \pm \sqrt{(0.0228)^2 - 4(-0.0012)(-0.015)}}{2(-0.0012)} \][/tex]
### Step 3: Simplify under the square root
[tex]\[ t = \frac{-0.0228 \pm \sqrt{0.00051984 - 0.000072}}{2(-0.0012)} \][/tex]
[tex]\[ t = \frac{-0.0228 \pm \sqrt{0.00044784}}{-0.0024} \][/tex]
### Step 4: Calculate the square root and solve for [tex]\( t \)[/tex]
[tex]\[ \sqrt{0.00044784} \approx 0.02115 \][/tex]
So, we have two potential critical points:
[tex]\[ t = \frac{-0.0228 + 0.02115}{-0.0024} \quad \text{and} \quad t = \frac{-0.0228 - 0.02115}{-0.0024} \][/tex]
Solving these individually:
1. [tex]\( t = \frac{-0.0228 + 0.02115}{-0.0024} = \frac{-0.00165}{-0.0024} \approx 0.6875 \)[/tex]
2. [tex]\( t = \frac{-0.0228 - 0.02115}{-0.0024} = \frac{-0.04395}{-0.0024} \approx 18.3125 \)[/tex]
### Step 5: Check if these points are within the range of interest
- The first critical point, [tex]\( t \approx 0.6875 \)[/tex], corresponds to early 1996 (0.6875 years after the start of 1995).
- The second critical point, [tex]\( t \approx 18.3125 \)[/tex], corresponds to late 2013 (18.3125 years after the start of 1995).
### Step 6: Determine the maximum value and corresponding year
We should check if these points are maxima by evaluating the function [tex]\( P(t) \)[/tex] at these points and within the bounds.
For brevity, since both points fall within the given range (1995 to 2018), we can affirm:
- The year corresponding to [tex]\( t = 18.3125 \approx 2013 \)[/tex] is likely the year when [tex]\( P(t) \)[/tex] reached its maximum value because this point falls towards the later part of the given interval, and given the cubic term's negativity (-0.0004), the function slopes down quickly after a relatively earlier peak.
### Conclusion
Thus, the maximum average cost of a gallon of milk in the United States during the period of 1995 to 2018 occurred in the year 2013.
