Optimization 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?

2 Answers

Relevance
  • Ramon
    Lv 7
    1 decade ago
    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

  • ?
    Lv 7
    1 decade ago

    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

Still have questions? Get your answers by asking now.