x = e^(8-p)
revenue is x*p = pe^(8-p)
profit = revenue-cost = xp-x = x(p-1) = (p-1)e^(8-p)
so, if y is the profit at price p,
dy/dp = (2-p)e^(8-p)
set the price at p=2
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?
Price =
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