Until recently, hamburgers at the city sports arena cost $2.60 each. The food concessionaire sold an average of 14 comma 000 hamburgers on game night. When the price was raised to $3.30, hamburger sales dropped off to an average of 7 comma 000 per night. Assuming a linear demand curve, find the price of a hamburger that will maximize the nightly hamburger revenue. If the concessionaire has a fixed cost of $1000 per night and the variable cost is $0.60 per hamburger, find the price of a hamburger that will maximize the nightly hamburger profit.

Since it’s assumed that the demand curve is linear, we can have the following expression representing the relationship between the demand and the pricing of a hamburger:

d = mp + b

where d is the demand (number of burgers sold), p is the pricing (USD). m and b are the slope and intersection of our linear equation, respectively. We can find the slope first:

[tex]m = \frac{d_2 – d_1}{p_2 – p_1} = \frac{7000 - 14000}{3.3 – 2.6} = \frac{-7000}{0.7}= -10000[/tex]

To solve for b we can plug in d = 7000, m = -10000 and p = 3.3

7000 = -10000*3.3 + b

b = 7000 + 3.3*10000 = 7000 + 33000 = 40000

Therefore, our equation for demand is d = -10000p + 40000

As it’s sold for p dollar for each burger, then the total revenue is

[tex]R = dp = (-10000p + 40000)p = -10000p^2 + 40000p[/tex]

To find the maximum value of this, we just need to take the 1st derivative and set it to 0

R’ = -20000p + 40000 = 0

20000p = 40000

p = 40000 / 20000 = 2 USD

If the concessionaire has a fixed cost of $1000 per night and the variable cost is $0.60 per hamburger. Our nightly cost would be:

C = 1000 + 0.6*d = 1000 + 0.6*(-10000p + 40000) = 1000 - 6000p + 24000 = 25000 – 6000p

Profit is revenue subtracted by cost

[tex]P = R – C = -10000p^2 + 40000p – (25000 – 6000p) = -10000p^2 +46000p – 25000[/tex]

Again to maximize the profit, we take the 1st derivative and set it to 0

P’ = -20000p + 46000 = 0

20000p = 46000

p = 46000 / 20000 = 2.3 USD