An analysis of the daily output of a factory assembly line shows that about 50t+t²- 1 tunits are produced after t hours of work, 0≤t≤8. What is the rate of production (in units per hour) when t = 4? 48 34 At t= 4, the rate of production is units per hour.

Calculus: Early Transcendentals
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Analysis of Factory Assembly Line Output**

An analysis of the daily output of a factory assembly line shows that about \( 50t + \frac{2}{48} t^3 \) units are produced after \( t \) hours of work, \( 0 \leq t \leq 8 \). 

**Question:**
What is the rate of production (in units per hour) when \( t = 4 \)?

**Solution:**

The rate of production is given by the derivative of the output function with respect to time \( t \). The output function is \( 50t + \frac{2}{48} t^3 \). To find the rate of production at \( t = 4 \), we first differentiate the function with respect to \( t \):

\[ \frac{d}{dt} \left( 50t + \frac{2}{48}t^3 \right) = 50 + \frac{2}{48} \cdot 3t^2 = 50 + \frac{1}{24} t^2 \]

Next, we substitute \( t = 4 \) into the derivative to find the rate of production:

\[ \text{Rate of production at } t = 4 = 50 + \frac{1}{24} (4)^2 = 50 + \frac{1}{24} \cdot 16 = 50 + \frac{16}{24} = 50 + \frac{2}{3} = 50.67 \text{ units per hour} \]

Thus, the rate of production at \( t = 4 \) is  \( \boxed{50.67} \) units per hour.
Transcribed Image Text:**Analysis of Factory Assembly Line Output** An analysis of the daily output of a factory assembly line shows that about \( 50t + \frac{2}{48} t^3 \) units are produced after \( t \) hours of work, \( 0 \leq t \leq 8 \). **Question:** What is the rate of production (in units per hour) when \( t = 4 \)? **Solution:** The rate of production is given by the derivative of the output function with respect to time \( t \). The output function is \( 50t + \frac{2}{48} t^3 \). To find the rate of production at \( t = 4 \), we first differentiate the function with respect to \( t \): \[ \frac{d}{dt} \left( 50t + \frac{2}{48}t^3 \right) = 50 + \frac{2}{48} \cdot 3t^2 = 50 + \frac{1}{24} t^2 \] Next, we substitute \( t = 4 \) into the derivative to find the rate of production: \[ \text{Rate of production at } t = 4 = 50 + \frac{1}{24} (4)^2 = 50 + \frac{1}{24} \cdot 16 = 50 + \frac{16}{24} = 50 + \frac{2}{3} = 50.67 \text{ units per hour} \] Thus, the rate of production at \( t = 4 \) is \( \boxed{50.67} \) units per hour.
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