A company runs food service concessions for sporting events throughout the country. Their marketing research department chose a particular football stadium to test market a new jumbo hot dog. It was found that the demand for the new hot dog is given approximately by p=8−ln(x),5≤x≤500, where x is the number of hot dogs (in thousands) that can be sold during one game at a price of p dollars. If the company pays 1 dollar for each hot dog, how should the hot dogs be priced to maximize the profit per game?

$3

Step-by-step explanation:

Given that:

p = 8 - ln(x) when 5 < x < 500

where;

x = The total number of dogs sold

Then;

The total revenue = x * p

R = x(8 - ln(x))

R = 8x - xln(x)

The Company thus pays 1 dollar per dog

i.e.

The total cost C = 1 * x = x

Then: Profit = R - C

P = 8x - xln(x) - x

P = 7x - xln(x)

Differentiating P in respect to x

dP/dx = 7 - d/dx(xln(x))

dP/dx = 7 - x*d/dx(ln(x)) - ln(x)*d/dx(x)

dP/dx = 7 - x(1/x) - ln(x)

dP/dx = 6 - ln(x)

Since this must be maximized, dP/dx is set to be equal to 0

6 - ln(x) = 0

ln(x) = 6

x = e^6

Now, p = 8 - ln(x)

Plug in the value of x :

p = 8 - ln(e^5)

p = 8 - 5

p = 3

Therefore, each dog must be priced at $3 to maximize the profit.

hi there!

first we need to find out how much they are adding to each number to make the next number.

ex: 4 + 4 = 8, 8 + 8 = 16, and so on.

so continuing with 32,

32 + 32 = 64

64 + 64 = 128

128 + 128 = 256

256 + 256 = 512

4, 8, 16, 32, 64, 128, 256, 512

the 8th term of the sequence will be 512

hope this ! ~

true

step-by-step explanation:

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