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.

To maximize revenue, set the price to $2

To maximize profit, set the price to $2.3

Step-by-step explanation:

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:

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

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

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

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11 -1> 1 -1+ a

-12 > a