Math Problem Statement

The number of visitors P to a website in a given week over a​ 1-year period is given by Upper P left parenthesis t right parenthesis equals 115 plus left parenthesis t minus 85 right parenthesis e Superscript 0.02 tP(t)=115+(t−85)e0.02t​, where t is the week and 1less than or equals≤tless than or equals≤52. ​a) Over what interval of time during the​ 1-year period is the number of visitors​ decreasing? ​b) Over what interval of time during the​ 1-year period is the number of visitors​ increasing? ​c) Find the critical​ point, and interpret its meaning. The critical point is left parenthesis 35 comma 14.312 right parenthesis(35,14.312).

Solution

We are given a function for the number of visitors to a website in a week over a 1-year period:

P(t)=115+(t85)e0.02tP(t) = 115 + (t - 85)e^{0.02t} where tt represents the week and 1t521 \leq t \leq 52.

Part (a) & (b) - Increasing and Decreasing Intervals:

To find where the number of visitors is increasing or decreasing, we need to analyze the first derivative of P(t)P(t), P(t)P'(t).

Step 1: Find the derivative P(t)P'(t)

We use the product and chain rules to differentiate P(t)P(t): P(t)=115+(t85)e0.02tP(t) = 115 + (t - 85)e^{0.02t} P(t)=ddt((t85)e0.02t)P'(t) = \frac{d}{dt}\left((t - 85)e^{0.02t}\right) Applying the product rule: P(t)=(t85)ddt(e0.02t)+e0.02tddt(t85)P'(t) = (t - 85) \cdot \frac{d}{dt}\left(e^{0.02t}\right) + e^{0.02t} \cdot \frac{d}{dt}(t - 85) Since ddt(e0.02t)=0.02e0.02t\frac{d}{dt}\left(e^{0.02t}\right) = 0.02e^{0.02t} and ddt(t85)=1\frac{d}{dt}(t - 85) = 1, we get: P(t)=(t85)0.02e0.02t+e0.02tP'(t) = (t - 85) \cdot 0.02e^{0.02t} + e^{0.02t} Factor out e0.02te^{0.02t}: P(t)=e0.02t(0.02(t85)+1)P'(t) = e^{0.02t}\left(0.02(t - 85) + 1\right) Simplify inside the parentheses: P(t)=e0.02t(0.02t1.7+1)P'(t) = e^{0.02t}(0.02t - 1.7 + 1) P(t)=e0.02t(0.02t0.7)P'(t) = e^{0.02t}(0.02t - 0.7)

Step 2: Analyze the sign of P(t)P'(t)

We now solve for when P(t)=0P'(t) = 0: e0.02t(0.02t0.7)=0e^{0.02t}(0.02t - 0.7) = 0 Since e0.02t0e^{0.02t} \neq 0 for any tt, we solve: 0.02t0.7=00.02t - 0.7 = 0 0.02t=0.70.02t = 0.7 t=0.70.02=35t = \frac{0.7}{0.02} = 35

Thus, the critical point occurs at t=35t = 35.

Step 3: Determine the intervals of increase and decrease

  • For t<35t < 35, 0.02t0.7<00.02t - 0.7 < 0, so P(t)<0P'(t) < 0 and the number of visitors is decreasing.
  • For t>35t > 35, 0.02t0.7>00.02t - 0.7 > 0, so P(t)>0P'(t) > 0 and the number of visitors is increasing.

Conclusion for parts (a) and (b):

  • The number of visitors is decreasing on the interval (1,35)(1, 35).
  • The number of visitors is increasing on the interval (35,52)(35, 52).

Part (c) - Critical Point:

The critical point is given as (35,14.312)(35, 14.312). This means:

  • At week t=35t = 35, the number of visitors reaches a minimum of approximately 14.312.
  • This critical point signifies the week during the year where the visitor count stops decreasing and starts increasing again.

Follow-up questions:

  1. How does the behavior of P(t)P(t) change as tt approaches the endpoints t=1t = 1 and t=52t = 52?
  2. What happens to the number of visitors as tt \to \infty, based on the nature of the exponential term?
  3. Can you find the total number of visitors throughout the year by integrating P(t)P(t)?
  4. How sensitive is the function to changes in tt, particularly near the critical point t=35t = 35?
  5. How would the intervals change if the growth rate in the exponential term was larger or smaller?

Tip: Critical points occur where the derivative equals zero or is undefined. Always check these points for potential maxima, minima, or points of inflection.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Critical Points
Exponential Functions

Formulas

P(t) = 115 + (t − 85)e^(0.02t)
Derivative of a product: (fg)' = f'g + fg'

Theorems

Critical Point Theorem
First Derivative Test

Suitable Grade Level

Grades 11-12 / Early College