Problem 62 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

At a certain factory, approximately \(q(t)=t^{2}+50 t\) units are manufac- tured during the first \(t\) hours of a production run, and the total cost of manufacturing \(q\) units is \(C(q)=0.1 q^{2}+10 q+400\) dollars. Find the rate at which the manufacturing cost is changing with respect to time 2 hours after production commences.

Step-by-Step Solution

Answer

The manufacturing cost is changing at a rate of 1,663.2 dollars per hour 2 hours after production commences.

Step 1: Identify the Given Functions

We are given two functions: the production function, which is approximated as \( q(t) = t^2 + 50t \) and the cost function, which is given by \( C(q) = 0.1q^2 + 10q + 400 \).

Step 2: Find the Production Rate

Differentiate the production function \( q(t) \) with respect to \( t \). This gives the rate at which units are manufactured: \( q'(t) = \frac{d}{dt}(t^2 + 50t) = 2t + 50 \).

Step 3: Calculate the Production Rate at \( t = 2 \)

Substitute \( t = 2 \) into the derivative: \( q'(2) = 2(2) + 50 = 4 + 50 = 54 \). So, units are being manufactured at a rate of 54 units per hour when \( t = 2 \).

Step 4: Express Cost as a Function of Time

Since \( q \) is a function of \( t \), the cost function also becomes a function of \( t \). We can denote this by \( C(t) = C(q(t)) \). To find \( \frac{dC}{dt} \), we need to apply the chain rule: \( \frac{dC}{dt} = \frac{dC}{dq} \cdot \frac{dq}{dt} \).

Step 5: Differentiate the Cost Function

Differentiate the cost function \( C(q) \) with respect to \( q \): \( \frac{dC}{dq} = \frac{d}{dq}(0.1q^2 + 10q + 400) = 0.2q + 10 \).

Step 6: Calculate \( \frac{dC}{dq} \) at \( q(t) \)

First, find \( q(2) \) by substituting \( t = 2 \) into \( q(t) \): \( q(2) = 2^2 + 50(2) = 4 + 100 = 104 \). Then substitute \( q = 104 \) into \( \frac{dC}{dq} \): \( \frac{dC}{dq} \bigg|_{q=104} = 0.2(104) + 10 = 20.8 + 10 = 30.8 \).

Step 7: Apply the Chain Rule

Now, use the chain rule to find \( \frac{dC}{dt} \): \( \frac{dC}{dt} = \frac{dC}{dq} \cdot \frac{dq}{dt} = 30.8 \cdot 54 \).

Step 8: Calculate the Final Rate of Change

Perform the multiplication: \( \frac{dC}{dt} = 30.8 \times 54 = 1,663.2 \).

Key Concepts

production function

In the context of manufacturing, the production function describes the relationship between the input of resources (like time, labor, and raw materials) and the output of finished products. In this exercise, the production function is defined as \( q(t) = t^2 + 50t \), where \( q(t) \) represents the number of units produced over time \( t \). This function helps us understand how the production rate changes over time.
By differentiating the production function, we can find the rate at which units are produced at any given time, \( t \). This differentiation provides us with \( q'(t) = 2t + 50 \), showing that the production rate depends on time and increases linearly.
For example, at \( t = 2 \) hours, the production rate is \( q'(2) = 54 \) units per hour, meaning that after 2 hours, the factory produces 54 units every hour.

cost function

The cost function represents the total cost of manufacturing units based on the number of units produced. In this exercise, we are given the cost function \( C(q) = 0.1q^2 + 10q + 400 \), where \( q \) is the number of units produced.
This quadratic function indicates that the cost increases not only linearly with the production but also quadratically, meaning that producing more units will significantly increase the total cost.
By knowing this cost function, we can determine the total cost for any quantity \( q \). In our specific case, we need to find how this cost changes with respect to time. To do this, we apply differentiation combined with the chain rule.

differentiation

Differentiation is a mathematical technique used to find the rate of change of a function. In this problem, we perform differentiation on both the production function and the cost function.
First, we differentiate the production function \( q(t) = t^2 + 50t \) with respect to time \( t \), which gives us the production rate \( q'(t) = 2t + 50 \).
Next, we differentiate the cost function \( C(q) = 0.1q^2 + 10q + 400 \) with respect to \( q \) to find \( \frac{dC}{dq} = 0.2q + 10 \).
These derivatives allow us to calculate how quickly units are produced and how the cost changes with each unit produced.

chain rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. In this exercise, the chain rule helps us find the rate of change of the cost function with respect to time.
Since the cost depends on the number of units \( q \), which in turn depends on time \( t \), we need to apply the chain rule:

• Find \( \frac{dq}{dt} \), the production rate from \( q(t) \). This was calculated as \( q'(t) = 2t + 50 \).
• Find \( \frac{dC}{dq} \), the rate of change of cost with respect to \( q \), which was calculated as \( 0.2q + 10 \).

Combining these using the chain rule, we get: ` \( \frac{dC}{dt} = \frac{dC}{dq} \times \frac{dq}{dt} \) 'Substituting the values at time \( t = 2 \):

• \( q(2) = 104 \)
• \( \frac{dC}{dq} \bigg|_{q=104} = 30.8 \)
• \( \frac{dq}{dt} \big|_{t=2} = 54 \)

Hence, the rate of change of the cost with respect to time at \( t = 2 \) is \( \frac{dC}{dt} = 30.8 \times 54 = 1,663.2 \). This tells us the cost increases by $1663.2 per hour at that specific time. Understanding this rate is crucial for managing the budget and making informed production decisions.