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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

Until recently hamburgers at the city sports arena cost 520 each The food concession air sold an average of 4500 hamburgers on game night when the price is rais class=

Respuesta :

Solution:

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

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

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

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