91Ó°ÊÓ

In this exercise we develop a model for the growth rate \(G\), in thousands of dollars per year, in sales of a product as a function of the sales level \(s\), in thousands of dollars. \({ }^{30}\) The model assumes that there is a limit to the total amount of sales that can be attained. In this situation we use the term unattained sales for the difference between this limit and the current sales level. For example, if we expect sales to grow to 3 thousand dollars in the long run, then \(3-s\) gives the unattained sales. The model states that the growth rate \(G\) is proportional to the product of the sales level \(s\) and the unattained sales. Assume that the constant of proportionality is \(0.3\) and that the sales grow to 2 thousand dollars in the long run. a. Find a formula for unattained sales. b. Write an equation that shows the proportionality relation for \(G\). c. On the basis of the equation from part b, make a graph of \(G\) as a function of \(s\). d. At what sales level is the growth rate as large as possible? e. What is the largest possible growth rate?

Step by step solution

01

Calculate unattained sales

The unattained sales represent the difference between the long-term sales goal and the current level of sales. Given that the sales are expected to grow to 2 thousand dollars in the long run, the formula for unattained sales is: \( u = 2 - s \).

02

Write the growth rate equation

According to the problem, the growth rate \( G \) is proportional to the product of the current sales level \( s \) and the unattained sales \( u \), where the constant of proportionality is 0.3. Using the formula for the unattained sales from Step 1, we get: \( G = 0.3 \, s \, (2 - s) \).

03

Graph the growth rate function

To graph \( G = 0.3 \, s \, (2 - s) \), notice that it's a quadratic equation opening downwards because the coefficient of \( s^2 \) is negative. Its graph will be a parabola with a maximum point. The graph of \( G \) as a function of \( s \) will show that \( G \) starts at zero, increases to a maximum, and then returns to zero as \( s \) approaches 2.

04

Determine the sales level for maximum growth rate

The growth rate will be maximum where the derivative \( \frac{dG}{ds} \) equals zero. Taking the derivative of \( G = 0.3 \, s \, (2 - s) \) gives: \( \frac{dG}{ds} = 0.3 \, (2 - 2s) \). Setting the derivative equal to zero: \( 2 - 2s = 0 \), solving for \( s \) gives \( s = 1 \).

05

Calculate the maximum growth rate

Substitute \( s = 1 \) back into the growth rate equation: \( G = 0.3 \, (1) \, (2 - 1) = 0.3 \, \times \, 1 \, \times \, 1 = 0.3 \). Thus, the largest possible growth rate is 0.3 thousand dollars per year.

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Key Concepts

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

Quadratic Functions

Quadratic functions are a type of polynomial where the highest-degree term is squared. They are typically written in the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These functions graph as parabolas, which can open upwards or downwards. The direction depends on the sign of \( a \):

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

In this exercise, the formula for the growth rate \( G = 0.3s(2-s) \) is a quadratic function. Here, the term \((2-s)\) can be expanded to create \(0.3(2s - s^2)\), resulting in a quadratic form. Since the coefficient of \(s^2\) is negative, this parabola opens downwards, indicating that the graph has a maximum point.

Proportionality

Proportionality occurs when two quantities increase or decrease in a constant ratio. When we say a quantity \( G \) is proportional to another quantity, it means \( G = k \cdot x \), where \( k \) is the constant of proportionality.
In our exercise, the growth rate \( G \) is proportional to the product of current sales level \( s \) and the unattained sales \( u \). Mathematically, this is expressed as:
\[G = 0.3 \cdot s \cdot (2 - s)\]

• This shows that as sales \( s \) increase, the growth rate \( G \) changes according to this proportional relationship.
• The constant of proportionality in this case is 0.3.

This means every unit increase in the product \( s \cdot (2-s) \) results in an increase of 0.3 times the same amount in \( G \).

Calculus Derivative

The calculus derivative is a tool used to determine the rate at which a function is changing at any given point, often used to find the maximum or minimum values of a function. When we want to find where the growth rate \( G \) is largest, we need to find the derivative of \( G \) with respect to \( s \).
For the given quadratic equation \( G = 0.3s(2-s) \), the derivative \( \frac{dG}{ds} \) can be computed using the power rule and the product rule:
\[\frac{dG}{ds} = 0.3(2 - 2s)\]

• To find the sales level where the growth rate is maximized, we set the derivative equal to zero: \( 2 - 2s = 0 \).
• Solving this gives \( s = 1 \).

This critical point \( s = 1 \) indicates the sales level where the growth rate \( G \) is maximized.

Graphing Functions

Graphing functions is a visual method of understanding the behavior of a function over a range of inputs. For quadratic functions, the graphs are parabolas.
To graph the growth rate function \( G = 0.3s(2-s) \), note that:

• The graph starts at \( s = 0 \), where \( G = 0 \).
• As \( s \) increases, \( G \) rises to its maximum at \( s = 1 \) and then decreases back to zero as \( s \) approaches 2.
• The graph is symmetrical around the maximum point due to the nature of parabolas.

This gives a clear, visual representation of how changing \( s \) affects \( G \). Using a graph, you can easily find maximum, minimum, and intercept points, helping to visualize theoretical predictions.