Until recently, hamburgers at the city sports arena cost $5.50 each. The food co

Question

Until recently, hamburgers at the city sports arena cost $5.50 each. The food concessionaire sold an average of 15,000 hamburgers on game night. When the price was raised to $5.90, hamburger sales dropped off to an average of 11,000 per night. (a) Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. (b) If the concessionaire had fixed costs of $1,500per night and the variable cost is S0.40 per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit. (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.) (b) If the concessionaire had fixed costs of $1,000 per night and the variable cost is S0.40 per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit The hamburger price that will maximize the nightly hamburger profit is

Explanation / Answer

(5.50 , 15000)
(5.90 , 11000)

x = price
y = no of hamburgers

Slope = (11000 - 15000)/(5.9 - 5.5) = -10000

Now, y = mx + c

y = -10000x + C

Using (5.5,15000) :
15000 = -10000(5.5) + C
So, C= 70000

Thus,
y = -10000x + 70000

Now, for max revenue, we have
x*y

R = x(-10000x + 70000)

R = -10000x^2 + 70000x

For max revenue,
dR/dx =0 :

-20000x + 70000 = 0

x = 3.5 ----> FIRST ANSWER

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b)
Total cost, C = 1000 + 0.4y
C = 1000 + 0.4(-10000x + 70000)
C = 1000 - 4000x + 28000
C = -3000x + 28000

Revenue - Cost is profit

P = -10000x^2 + 70000x - (-3000x + 28000)

P = -10000x^2 + 73000x - 28000

Now, for max profit,
dP/dx = 0 :

-20000x + 73000 = 0

x = 73000/20000

x = 3.65 dollars ---> ANS

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