Optimization question?

Question

The demand for the new hot dog is given approximately by:

p = 8−ln(x), 5 < equal to x < equal to 500

x = # of hot dogs (in thousands) that can be sold during one game at a price of p dollars

It cost $1 for each hot dog

How should the hot dogs be priced to maximize the profit per game?

Answer 1

Favorite Answer

the profit equation is

Prof=(p-1)x

=[8-ln(x)-1]x

=7x-xln(x)

Prof`=7-ln(x)-1 to maximize the profit we make Prof'=0 so

0=6 -ln(x)

ln(x)=6

x=e^(6)

x=403,429

p=8-ln(403.429)

p=$2

Answer 2

R(x) = p*x

R(x) = 8*x - x*ln(x)

C(x) = 1*x = x

P(x) = 8*x - x*ln(x) - x

P(x) = 7*x - x*ln(x)

diff(7*x-x*ln(x), x) = 6 - ln(x)

solve(6 - ln(x))

x = exp(6.)

x = 403.4287935

p = 8 − ln(x)

p = 8 - ln(403.4287935)

p = 2