Question

A model for​ consumers' response to advertising is given by the equation N(a)=2600 + 470ln (a) Where​ N(a) is the number of units​ sold, a is the amount spent on​ advertising, in thousands of​ dollars, & a≥1.

Required:
a. How many units were sold after spending ​$1,000 on​ advertising?
b. Find N′(a).
c. Find the maximum​ value, if it exists.
d. Find lim a→[infinity] N′(a).

Answer 1

a.
`N(1)=2600`

b.
`N'(a) = 470/a`

c. N(a) has no maximum value, max N'(a) = 470 (when a = 1)

d. lim a→[infinity] N′(a) = 0

Explanation:

a.

the variable 'a' is the amount spent in thousands of dollars, so $1,000 is equivalent to a = 1. Then, we have that:

`N(1)=2600 + 470ln(1)`

`N(1)=2600 + 470*0`

`N(1)=2600`

b.

To find the derivative of N(a), we need to know that the derivative of ln(x) is equal (1/x), and the derivative of a constant is zero. Then, we have:

`N'(a) = 2600' + (470ln(a))'`

`N'(a) = 0 + 470*(1/a)`

`N'(a) = 470/a`

c.

The value of 'ln(a)' increases as the value of 'a' increases from 1 to infinity, so there isn't a maximum value for N(a).

The maximum value of N'(a) is when the value of a is the lower possible, because 'a' is in the denominator, so the maximum value of N'(a) is 470, when a = 1.

d.

When the value of 'a' increases, the fraction '470/a' decreases towards zero, so the limit of N'(a) when 'a' tends to infinity is zero.