Quadratic Relations (ex. Graphing)

Question

The captain of a riverboat cruise charges $36 per person, including lunch. The cruise averages 300 customers a day. The captain is considering increasing the price. A survey of customers indicates that for every $2 increase in price, there would be 10 fewer customers.

Write and solve a quadratic model to find the price increase that would maximize the revenue.

Answer 1

At $36, 300 customers

raise price $2, lose 10 customers

two points on the line are (300,36) and (290, 38)

y=mx+b

P=mC+b

slope=m=(36-38)/(300-290) = -2/10= -1/5

P=-C/5+ b

36=-300/5 +b

36=-60+b

b=36+60= 96

b=96

P=-C/5 +96

maximum revenue

R=CP= -C^2/5 +96C

take the derivative and set equal to zero

R' = -2C/5 +96 = 0

2C -480= 0

C=240

P= $48

max R= 48x240= $11520

or

graph the equation

the revenue maximizing point is the midpoint of the line segment

(240,48)

to maximize revenue increase price $12, from $36 to $48