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Part 1

The relationship between revenue generated and the number of employees for a particular company can be approximated by 

77nminus−1010Requals=5353​, 

where R is revenue in billions of dollars and n is the number of employees in thousands. Find the rate of change of employees with respect to​ time, assuming that revenue was increasing at a rate of 

​$.5656 

billion per year.

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Part 1

Differentiate both sides of the equation with respect to time.

enter your response hereStartFraction dn Over dt EndFractiondndtminus−enter your response hereStartFraction dR Over dt EndFractiondRdtequals=enter your response here

Part 2

Find the rate of change of the number of employees with respect to time. Select the correct choice and fill in the answer box to complete your choice.

​(Round to the nearest whole number as​ needed.) 

A.

The number of employees is decreasing by 

enter your response here 

employees per year. 

B.

The number of employees is increasing by 

enter your response here 

employees per year. 

To solve this problem, we will first differentiate the given equation with respect to time $ t $ and then solve for the rate of change of employees with respect to time $ \frac{dn}{dt} $.

### Part 1: Differentiate both sides of the equation with respect to time.

Given equation:
\[ 77n - 10R = 53 \]

Differentiating both sides with respect to $ t $:

\[ 77 \frac{dn}{dt} - 10 \frac{dR}{dt} = 0 \]

We rearrange the differentiated equation to solve for $ \frac{dn}{dt} $:

\[ 77 \frac{dn}{dt} = 10 \frac{dR}{dt} \]

\[ \frac{dn}{dt} = \frac{10}{77} \frac{dR}{dt} \]

Now, we substitute the given rate of change of revenue $ \frac{dR}{dt} = 0.56 $ billion per year:

\[ \frac{dn}{dt} = \frac{10}{77} \times 0.56 \]

### Part 2: Calculate the rate of change of the number of employees with respect to time.

\[ \frac{dn}{dt} = \frac{10 \times 0.56}{77} \]

\[ \frac{dn}{dt} = \frac{5.6}{77} \]

\[ \frac{dn}{dt} \approx 0.0727 \]

So, the rate of change of the number of employees with respect to time is approximately $ 0.0727 $ thousand employees per year. To find the number of employees per year, we multiply by 1000:

\[ \frac{dn}{dt} \approx 0.0727 \times 1000 \]

\[ \frac{dn}{dt} \approx 73 \]

Since the result is positive, the number of employees is increasing by 73 employees per year.

### Answer:

B. The number of employees is increasing by $ 73 $ employees per year.

Would you like any further details or have any questions about the solution?

### Related Questions:
1. How would the result change if the revenue was increasing at a different rate?
2. What is the interpretation of the negative coefficient in the original equation?
3. How can we determine the number of employees if we know the revenue?
4. What happens to the number of employees if the revenue is constant?
5. Can this model be used to predict future revenue? Why or why not?
6. What assumptions are we making about the relationship between revenue and employees?
7. How would you solve the problem if the relationship was non-linear?
8. What other factors might affect the number of employees besides revenue?

### Tip:
When differentiating implicit functions, remember to apply the chain rule and consider each variable's rate of change with respect to time.

Explore the rate of change of employees with respect to revenue and time using differentiation techniques. Solve a linear equation relating revenue and number of employees, considering a given rate of revenue increase. Suitable for advanced high school students and beyond.

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