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).

a.

b.

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

d. lim a→[infinity] N′(a) = 0

Step-by-step 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:

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:

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.

570, the answer is 570, hope it

twenty-six miles over two gallons because the proportion for the original 13 miles to one gallon would be written 13/1 and, doing the same thing to the top and bottom, i multiplied by 2 and got 26/2