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Chapter 3: Problem 24

Milk Storage The highest temperature at which milk can be stored to remain fresh for \(x\) days can be modeled as $$ f(x)=-9.9 \ln x+60.5^{\circ} \mathrm{F} $$ (Source: Simplified model based on data from the back of a milk carton from Model Dairy) a. What is the highest temperature at which milk can be stored to remain fresh for at least 5 days? How quickly is the required temperature changing at this point? b. What is the average rate of change in required storage temperature between 3 and 7 days? c. As the number of days increases, does the rate of change of temperature increase or decrease?

Short Answer

Expert verified
a) 48.1掳F; -1.98掳F/day. b) -2.055掳F/day. c) The rate of change decreases as days increase.

Step by step solution

01

Substitute x=5 into the function

To find the highest temperature for 5 days, substitute \(x = 5\) into the function \(f(x) = -9.9 \ln x + 60.5\). Calculate \(\ln 5\), then compute the value of \(f(5)\).
02

Differentiate the function

To find how quickly the required temperature is changing when \(x=5\), differentiate the function \(f(x)\) with respect to \(x\). The derivative is \(f'(x) = -9.9 \cdot \frac{1}{x}\).
03

Evaluate the derivative at x=5

Evaluate the derivative \(f'(x) = -9.9 \cdot \frac{1}{x}\) at \(x = 5\) to find the rate of change of the required temperature at that point.
04

Calculate the average rate of change between x=3 and x=7

Use the average rate of change formula: \(\frac{f(7) - f(3)}{7-3}\) where \(f(x) = -9.9 \ln x + 60.5\). Compute \(f(3)\) and \(f(7)\) then find the difference and divide by 4.
05

Analyze the derivative function f'(x)

Determine the behavior of \(f'(x) = -9.9 \cdot \frac{1}{x}\) as \(x\) increases. Since \(\frac{1}{x}\) decreases as \(x\) increases, \(f'(x)\) becomes less negative, meaning the rate of change decreases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Function
In this exercise, the temperature function is given as a logarithmic equation: \[ f(x) = -9.9 \ln x + 60.5 \text{掳F} \]This function models the highest temperature milk can be stored at for it to remain fresh over a period of \(x\) days.
The higher the value of \(x\), the lower the temperature it should be stored at, to keep the milk fresh.
This inversely proportional relationship implies that as the duration of storage increases, lower temperatures are required.
The 60.5 represents the base temperature when no days are involved. The \(-9.9\) term scales down the temperature based on the logarithm of the number of days.
Logarithmic Functions
Logarithmic functions like \(f(x) = -9.9 \ln x + 60.5\) are unique because they include a logarithmic term. The logarithm, specifically the natural logarithm (ln) in this context, is used to transform rapidly increasing numbers into something more manageable.
By using a logarithm, we can depict growth that slows down as \(x\) increases. This is suited for modeling situations where growth rates decrease over time.
When computing \(\ln\) values, remember:
  • \(\ln(1) = 0\)
  • \(\ln(e) = 1\)
  • Values for other numbers involving base \(e\) can be computed using a scientific calculator.
Differentiation
Differentiation involves finding the rate at which a function changes as its variables change.
For the temperature function \(f(x) = -9.9 \ln x + 60.5\), we need to differentiate to understand how fast the required storage temperature changes per day.
The derivative \(f'(x) = -9.9 \cdot \frac{1}{x}\) represents this rate of change.
When \(x = 5\), for instance, substituting this into the derivative gives us the rate of change of temperature at that day. This informs us how much cooler the storage needs to be as days pass.
Differentiation is a crucial tool in calculus for analyzing trends and patterns over continuous data.
Average Rate of Change
The average rate of change is useful to determine how much a function's output changes over an interval.
For instance, we use this concept to find how much the required temperature changes between 3 and 7 days.This is done using the formula: \[\frac{f(7) - f(3)}{7-3}\]
This result offers a simple view of the overall trend across the interval.
In this exercise, it provides perspective on how dramatically the storage temperature requirements decline over those specific days.
  • First, calculate the function values at the beginning: \(f(3)\)
  • Next, at the end: \(f(7)\)
  • Finally, divide their difference by the length of the interval, 4 days in this case.

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Most popular questions from this chapter

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