Until recenty, hamburgers at the city sports arena cost $4.20 each. The food con

Question

Until recenty, hamburgers at the city sports arena cost $4.20 each. The food concessionaire sold an average of 9,000 hamburgers on game night. When the price was raised to $4.90, hamburger sales dropped off to an average of 5,500 per night (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue 50.70 per hamburger, find the price of a hamburger th hat will maximize the nightly hamburger profit (b) If the concessionaire had fixed costs of 51,500 per night and the variable cost is (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue The hamburger price that will maximize the nightly hamburger revenue is S ( Round to the nearest cent as needed.)

Explanation / Answer

(a)

To assume linear demand, an equation between price (p) and no. of hamburgers (H) sold will be established.

H = m*p + constant

points = (p,H) = (4.2,9000) and (4.9, 5500)

m = slope = (9000-5500)/(4.2-4.7) = -3500/0.7 = -5000 burgers per dollar

H = -5000p + constant

9000 = -5000*4.2 + constant

constant = 9000 + 5000*4.2 = 30000

so equation is H = -5000p + 30000

Revenue = price * hamburgers = (-5000p+30000) * p

R = -5000p^2 -30000p

Revenue will be maximum where first derivate of R will be 0.

dR/dp = -10000p +30000 = 0

p = $3

Revenue max = 3*(-5000*3 + 30000) = 15000*3 = 45000

please upvote the answer.

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