Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

A business is prospering in such a way that its total (accumulated) profit after \(t\) years is \(1000 t^{2}\) dollars. (a) How much did the business make during the third year (between \(t=2\) and \(t=3\) )? (b) What was its average rate of profit during the first half of the third year, between \(t=2\) and \(t=2.5\) ? (The rate will be in dollars per year.) (c) What was its instantaneous rate of profit at \(t=2\) ?

Step-by-Step Solution

Answer

(a) 5000 dollars. (b) 4500 dollars per year. (c) 4000 dollars per year.

Step 1: Calculate Total Profit at t=3

The profit after 3 years is calculated using the given formula for total accumulated profit:\[ P(3) = 1000 \times 3^2 = 1000 \times 9 = 9000 \text{ dollars} \]

Step 2: Calculate Total Profit at t=2

The profit after 2 years is calculated using the given formula for total accumulated profit:\[ P(2) = 1000 \times 2^2 = 1000 \times 4 = 4000 \text{ dollars} \]

Step 3: Find Profit During the Third Year

The profit made during the third year is the difference between the total profit after 3 years and after 2 years:\[ P(3) - P(2) = 9000 - 4000 = 5000 \text{ dollars} \]

Step 4: Calculate Total Profit at t=2.5

The profit after 2.5 years is calculated using the given formula for total accumulated profit:\[ P(2.5) = 1000 \times (2.5)^2 = 1000 \times 6.25 = 6250 \text{ dollars} \]

Step 5: Calculate Profit for the First Half of the Third Year

The profit during the first half of the third year is the difference between the total profit after 2.5 years and after 2 years:\[ P(2.5) - P(2) = 6250 - 4000 = 2250 \text{ dollars} \]

Step 6: Calculate Average Rate of Profit in First Half of Third Year

The average rate of profit during the interval from \(t=2\) to \(t=2.5\) can be calculated as follows:\[ \text{Average Rate} = \frac{\text{Profit over interval}}{\text{Time interval}} = \frac{2250}{0.5} = 4500 \text{ dollars per year} \]

Step 7: Differentiate the Profit Function

To find the instantaneous rate of profit, we differentiate the profit function with respect to time \(t\):\[ \frac{dP}{dt} = \frac{d}{dt}(1000t^2) = 2000t \]

Step 8: Calculate Instantaneous Rate at t=2

Substitute \(t = 2\) into the differentiated function to find the instantaneous profit rate at \(t=2\):\[ \frac{dP}{dt} \bigg|_{t=2} = 2000 \times 2 = 4000 \text{ dollars per year} \]

Key Concepts

Accumulated Profit Formula

To understand how much profit a business accumulates over time, we use the accumulated profit formula. In this exercise, the formula for the total accumulated profit after \(t\) years is given by \(1000t^2\) dollars. This formula helps to determine the profit at any point in time.

• For instance, if you want to know the profit after 3 years, you simply substitute \(t=3\) into the formula: \(P(3) = 1000 \times 3^2 = 9000\) dollars.
• To find the profit after 2 years, substitute \(t=2\): \(P(2) = 1000 \times 2^2 = 4000\) dollars.

Using this formula, we can easily calculate the profit at specific time intervals and understand how the profit grows as time progresses.

Average Rate of Change

The average rate of change gives insight into how quickly profit is increasing over a specific time period. It tells us the average profit rate per year.

• For the time interval between years 2 and 2.5, we calculate the profit difference and divide it by the length of the interval.

The profit at \(t=2.5\) is \(6250\) dollars, while at \(t=2\) it's \(4000\) dollars. Thus, the profit over the interval is \(2250\) dollars.To find the average rate over the interval, divide this profit by the time span \(0.5\) years:\[ \text{Average Rate} = \frac{2250}{0.5} = 4500 \text{ dollars per year} \] This calculation gives a clearer picture of the business's growth pace during that period.

Instantaneous Rate of Change

The instantaneous rate of change is like a snapshot of the profit rate at a precise moment, unlike the average rate which is over a period.To find this rate, we use calculus to differentiate the profit function with respect to time \(t\).

• The derivative of \(1000t^2\) is calculated as \(2000t\).

This derivative, \(\frac{dP}{dt} = 2000t\), represents our instantaneous rate of profit at any given time \(t\). To find the instantaneous rate at \(t=2\), substitute into the derivative: \[ \frac{dP}{dt} \bigg|_{t=2} = 2000 \times 2 = 4000 \text{ dollars per year} \]This tells us that at exactly 2 years, the rate at which profit is increasing is 4000 dollars per year.

Differentiation in Calculus

Differentiation in calculus is a powerful tool used to calculate rates of change, such as profit rates in business.This process involves finding the derivative of a function, which provides a formula for the rate of change at any given point.

• In this exercise, the profit function \(1000t^2\) is differentiated to get \(2000t\),

which tells us how profit changes over time.Differentiation gives us the ability to make strategic decisions based on instantaneous data, such as reallocating resources when instantaneous profit rates aren't favorable.