Question 1172717
A city centre tour guide currently charges $34 for a full day’s tour. The average number of customers is 48. 
Market research suggests that for every $1 increase in tour price, the guide can
expect to lose two customers per tour.
(a) Show that if the price increase is $x, then the expected revenue from each tour is {{{-2x^2-20x+1632}}}
:
Revenue = price * no. of customers
R(x) = (34+x)*(48-2x)
FOIL
R(x) = 1632 - 68x + 48x + -2x^2
Combine like terms. Arrange as a quadratic
R(x) = -2x^2 - 20x + 1632
:
:
(b) The guide needs to ensure that the expected revenue is at least $1440. By solving a quadratic inequality, find the range of prices that need to be charged.
:
-2x^2 - 20x + 1632 >= 1440
-2x^2 - 20x + 1632 - 1440 >= 0
-2x^2 - 20x + 192 >= 0
simplify, divide by -2
x^2 + 10x - 96 <= 0, negative mult, reverses inequality
Factors to
(x+16)(x-6) <= 0
positive solution
x <= 6 (a price increase from 34)
x >= -16 (a price drop from 34)
price range from $40 (34+6)
(34+6)(48-12) = 1440
and
to $18 (34-16)
(32-16)(48+32) = 1440
Price range $18 to $40 to make at least $1440
:
:
(c) What price should be charged to maximize expected revenue?
This will occur on the axis of symmetry where a=-2, b =-20
x = {{{(-(-20))/2(-2)}}}
x = -5
(34-5)(48-2(-5)) = 
29 * (48+10) = $1682 is max revenue

Graphically is will make more sense. Green line is $1440
{{{ graph( 300, 200, -24, 24, -500, 2000, -2x^2-20x+1682, 1440) }}}