Question

Until recently, hamburgers at the city sports arena cost $5.20 each. The food concession air sold an average of 4500 hamburgers on game night. when the price is raised to $5.90, hamburger sales dropped off to an average of 2750 per nighta. Assuming a linear demand curve, find the price of hamburger that will maximize nightly hamburger revenue. The hamburger place that will maximize the nightly revenue is $ (round to the nearest cent as needed) b. if the concessionaire had fixed cost of $2500 per night and the variable cost is $.50 per night, find the price of a hamburger that will maximize nightly hamburger profit

Answer 1

From the question,

When price = $5.2, demand = 4500

When price = $5.9, demand = 2750

Assuming a linear demand curve,

We will have to determine the demand function (which is a linear function)

Using the knowledge of equation of a straight line given two points, we can determine the demand function to be

I.e when x = 5.2, y=4500

When x = 5.9, y = 2750

Thus, the equation is y = -2500x + 17500

In Demand function, we have

d = -2500p + 17500

Thus, the demand function is

d = -2500p + 17500

Revenue function = demand function x price

`\begin{gathered} Thus,\text{ } \\ R=(-2500p+17500)\text{p} \\ R=-2500p^2+17500p \\ \end{gathered}`

To determine the price that maximize the nightly hamburger revenue, plot the graph of Revenue function above and determine the value of p at the maximum point

The graph of Revenue function is shown below

From the graph, at maximum point, p = 3.5

Thus, the price that maximize the nightly hamburger revenue is $3.5