# A car dealership is trying to determine how much to spend on advertising during the upcoming...

## Question:

A car dealership is trying to determine how much to spend on advertising during the upcoming fiscal year.

Let {eq}N(x) {/eq} be the expected number of cars that will be sold during the year if the dealer spends {eq}x {/eq} thousand dollars on advertising. Market research indicates annual sales will change at a rate of

{eq}\displaystyle\; N'(x) = \frac{30}{\sqrt{1 + 0.2x}} {/eq}

during the upcoming fiscal year.

(a) Determine the expected increase in sales if the annual advertising budget is increased from {eq}$40\textrm{,}000 {/eq} to {eq}$45\textrm{,}000 {/eq}.

(b) Determine the expected increase in sales if the annual advertising budget is increased from {eq}$50\textrm{,}000 {/eq} to {eq}$55\textrm{,}000 {/eq}.

(c) In which of (a) or (b) does the {eq}$5\textrm{,}000 {/eq} increase in annual advertising expenditures lead to a greater increase in auto sales? Explain.

(d) Use what you learned from (a) through (c) to determine whether {eq}N(x) {/eq} is concave up or concave down and explain your reasoning.

## Finding Increase in Sales Upon Increasing Advertising Budget

The rate of change of sales in a car dealership is given as an explicit function of the number of dollars, in thousands, spent by the car dealership in advertising. Using a definite integral that uses the Fundamental Theorem of Calculus to evaluate, we find the increase in the sales of cars when the advertising amount is increased from a given quantity to another. Then using the same concept we find the increase in car sales for another different increase in the advertising budget.

## Answer and Explanation:

(a) The increase in sales in this part will be given by the following definite integral:

{eq}\displaystyle Increase = N(45)-N(40)=\int_{40}^{45} N'(x) \; dx = \int_{40}^{45} \frac {30}{\sqrt {1+0.2x}} \; dx \qquad (1) {/eq}

The integral in (1) above is evaluated as follows:

{eq}\displaystyle \int_{40}^{45} \frac {30}{\sqrt {1+0.2x}} \; dx = \left[ \frac {30}{0.2} \; \frac {(1+0.2x)^{-1/2+1}}{-1/2+1} \right]_{40}^{45} = 300 \left[ (1+0.2(45))^{1/2}-(1+0.2(40))^{1/2} \right] = 48.69 \approx 49. {/eq}

**The increase in the numbers of sales of cars here is about 49.**

(b) Just like in part (a), {eq}\displaystyle Increase = N(55)-N(50)=\int_{50}^{55} N'(x) \; dx = \int_{50}^{55} \frac {30}{\sqrt {1+0.2x}} \; dx \qquad (2) {/eq}

The integral in (2) above is evaluated as follows:

{eq}\displaystyle \int_{50}^{55} \frac {30}{\sqrt {1+0.2x}} \; dx = \left[ \frac {30}{0.2} \; \frac {(1+0.2x)^{-1/2+1}}{-1/2+1} \right]_{50}^{55} = 300 \left[ (1+0.2(55))^{1/2}-(1+0.2(50))^{1/2} \right] = 44.24 \approx 44. {/eq}

**The increase in the numbers of sales of cars here is about 44.**

(c) In part (a) there is a greater increase in the number of cars sold when the advertising amount is increased from $40,000 to $45,000.

(d) It is concave down since the amount of change in the value of N(x) is decreasing as the value of x gets bigger. That means the derivative of N(x) or N'(x) is decreasing with x increasing. This implies the second derivative is negative which indicates downwards concavity. Eventually the change will become zero.

#### Learn more about this topic:

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2