Question

A national food service runs concession

stands at sporting events. Its marketing re-
search department has found that at a price
of $p per pizza the demand for pizzas is given
approximately by
p = 35-7 ln x
where x is the number of pizzas sold. If
the concessionaire pays $12 for each pizza,
estimate the maximum profit, Pmax, the food
service can make at a sporting event.

Answer 1

Answer:
Therefore, the maximum profit the food service can make at a sporting event is approximately $219.06.

Explanation:

To find the maximum profit, we need to determine the price that will maximize revenue, and then subtract the cost to get the profit.

First, we need to find the revenue function. Revenue is the price per pizza multiplied by the quantity sold, or:

R(x) = p(x) * x

where p(x) is the price function given by:

p(x) = 35 - 7 ln(x)

Substituting this into the revenue function, we get:

R(x) = (35 - 7 ln(x)) * x

To find the maximum revenue, we can take the derivative of R(x) with respect to x and set it equal to zero:

dR/dx = 35 - 7 ln(x) - 7 = 0

Solving for x, we get:

ln(x) = (35 - 7)/(-7) = -4

x = e^(-4) ≈ 0.0183

So the maximum revenue is obtained by selling approximately 0.0183 * 1000 = 18.3 pizzas (assuming a demand of 1000 pizzas).

To find the price that maximizes revenue, we can substitute x = 18.3 into the price function:

p(18.3) = 35 - 7 ln(18.3) ≈ $24.09

So the concessionaire should charge about $24.09 per pizza to maximize revenue.

Now we can find the maximum profit by subtracting the cost per pizza from the price:

Pmax = (24.09 - 12) * 18.3 ≈ $219.06