Revenue Problems
1. You run a canoe-rental business on a small river in Ohio. You currently charge $12 per canoe and
average 36 rentals a day. An industry journal says that for every 50 cent increase in rental price, the
average business can expect to lose two rentals a day. Use this info to attempt to maximize your
income. What should you charge?
Revenue = Charge x Number = $12 x 36 = $432
Changing Revenue: Let x represent each change in charge and number.
Steps:
 For each change, the charge goes up by 50 cents
(12  .50 x) and the number of rentals goes down by 2
rentals (36  2 x )
 Revenue = Charge x Number of Rentals
 Simplify the quadratic.
R  (12  0.50 x)(36  2 x)
 432  24 x  18 x  x 2
 432  6 x  x 2
b
2a
6

2(1)
 3
x
You are trying to find the maximum so use the vertex
formula.
The negative answer suggests that the maximum will be
achieved when lowering the price by $.50 three times.
The charge should be
$12  0.50(3)
:  $12  $1.50
 $10.50
The Maximum Revenue will be
$12  0.50(3)36  2(3)
 $10.50  42
 $ 441
2. When priced at $30 each, a toy has annual sales of 4000 units. The manufacturer estimates that each
$1 increase in cost will decrease sales by 100 units. Find the price that will maximize total revenue. At
which price will revenue be zero?
Steps:
R  (30  1x)( 4000  100 x)
 120000  3000 x  4000 x  100 x
 120000  1000 x  100 x 2
b
2a
1000

2(100)

5
x 
So increasing the price one half the dollar amount will
maximize the revenue. The price should be $30.50
The factors are
x  400
 For each change, the charge goes up by 50 cents
(12  .50 x) and the number of rentals goes down by 2
rentals (36  2 x )
 Revenue = Charge x Number of Rentals
 Simplify the quadratic.
You are trying to find the maximum so use the vertex
formula.
(400  x)(300  x)  0
: So 400  x  0
2
or 300  x  0
x   300
You are trying to find the roots so factor the equation
and solve for x.
So if you drop the price $300, the toys cost $0 so no
revenue. If you use 400, the second factor will equal
4000 so you will have no sales and no revenue.
Number Problems
3.
Let x represent one number.
Then the other number is 19-x
You are looking for the value of x and 19 - x, so you
are solving for x.
Pr oduct  x(19  x)
x 2  19 x  60  0
 19 x  x 2
( x  15)( x  4)  0
19 x  x 2  60
x  15
x 2  19 x  60  0
or x  4
So the two numbers are 15 and 4.
4.
Let x be one number and 3x be the other.
Equation:
3 x 2  4 x  55  0
Product = Sum + 55
x 3x   x  3x  55
3x 2
You are looking for the value of x and 3x, so you are
solving for x.
P= -165
&
S = -4
3 x 2  15 x  11x  55  0
3 x( x  5)  11( x  5)  0
 4 x  55
( x  5)(3 x  11)  0
3x 2  4 x  55  0
x5
or x  
11
3
So the two numbers are 5 and -11/3.
Geometry Problems
5.
1
bh
2
1
  x  x  7 
2
1
 x 2  7 x)
2
A
x
x+7
1 2
( x  7 x)
2
588  x 2  7 x
294 
x 2  7 x  588  0


Draw the triangle and label its sides.
Use the Area Formula to set up the equation.
Set the Area equation to the given area.
You are solving for x, so factor or use the Q. F.
x 2  7 x  588  0
( x  21 )x  28   0
So x  21 or x  28
inadmissib le
Motion Problems
6.
Set up the table with the info.
Let x represent the speed of the car
D
V
Car
600
x
Truck
600
x - 20
Use two of the columns to fill in the third column.
T
600
x
600
longer time
x  20
Set up an equation with the third column.
600
600

5
x  20
x
600 x  600( x  20)  ( x)( x  20)(5)
Diff = 5
600 x  600 x  12000  ( x 2  20 x)5
Use these two columns to fill in the third, using the formula
for time.
12000

5 x 2  100 x
5 x 2  100 x  12000  0
x 2  20 x  2400  0
( x  60 )( x  40 )  0
So x  60 or  40
inadmissib le
7.
Let x represent the rate of the man rowing in still
water.
D
V
Downstream
8
x+1
Upstream
8
x-1
T
8
x 1
8
x 1
Total = 6
Set up the table with the info.
Use two of the columns to fill in the third column.
Set up an equation with the third column.
8
8

6
x 1
x 1
8 x  8  8 x  8  6( x  1)( x  1)
16 x  6( x 2  1)
16 x  6 x 2  6
6 x 2  16 x  6  0
Use these two columns to fill in the third, using the formula
for time.
(3 x  1)( x  3)  0
1
3
inadmissib le
So x  
or 3