Until? recently, hamburgers at the city sports arena cost ?$5.80 each. The food

Question

Until? recently, hamburgers at the city sports arena cost ?$5.80 each. The food concessionaire sold an average of 22,000 hamburgers on game night. When the price was raised to ?$6.3?0, hamburger sales dropped off to an average of 17,000 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 ?$2 ,000 per night and the variable cost is

?$0.70 per? hamburger, find the price of a hamburger that will maximize the nightly hamburger profit?

Please explain process to solve this, very confused with ALL the steps. Will rate :)

Explanation / Answer

A)

Given two points are (22000, 5.80) and (17000, 6.30)

Find slope = (6.30 -5.80)/(17000 -22000) =  -0.0001

Write,  linear demand? equation as

p - 5.80 =   -0.0001 (x-22000)

p =   -0.0001 x +8

Find revenue as

R=p*x = -0.0001 x² +8x

Now to maximize find derivative and set to 0 as

-0.0002x +8 =0

0.0002x =8

x= 8/0.0002

x =40000

Hence

p = -0.0001 *40000 +8 =4

Thus, price of a hamburger that will maximize the nightly hamburger profit is $4.

b)

fixed costs of ?$2 ,000 per night and the variable cost is ?$0.70 per? hamburger, hence find cost function as

C(x) = 2000 +0.70x

Find Profit as

P(x)=R(x)-C(x)

=(-0.0001 x² +8x) -(2000 +0.70x)

=-0.0001 x^2 + 7.3 x - 2000

Now to maximize find derivative and set to 0 as

-0.0002x+7.3=0

x=7.3/0.0002

x=36500

hence

p =   -0.0001 *36500 +8 =$4.35

Hence, price of a hamburger that will maximize the nightly hamburger profit is $4.35.

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