Alg II - Chapter 5 Word Problems
Pg 253 #54: The woodland jumping mouse can hop
surprisingly long distances given its small size. A
relatively long hop can be modeled by: y = -2/9x(x-6)
where x and y are measured in feet.
a) How far can a woodland jumping mouse hop?
Solution: When it says “how far,” it’s really asking
for the distance between the x-intercepts. Why?
So how far?
Well, it’s
pretty clear
that it can
jump six feet.
(Horizontal
distance)
b) How high can it hop?
Solution: Whenever you’re asked, “how high,”
it’s really asking for the y-value of the vertex. For an
upside-down parabola, the vertex is guaranteed to
be the highest point of the parabola.
For this graph, the vertex is at (3, 2).
So the mouse can jump a maximum of 2 feet high.
Alg II - Chapter 5 Word Problems
Pg 261 #90:
You have made a quilt that is 4 feet by 5 feet. You
want to use the remaining 10 square feet of fabric to
add a decorative border of uniform width. What
should the width of the border be?
Solution: Draw a picture!
Note: Uniform = same.
x
Total Area = (5 + 2x)(4 + 2x)
20 + 10 = (5 + 2x)(4 + 2x)
x
5 30
= (5 + 2x)(4 + 2x)
30
= 20 + 4x2 + 18x
4
0
= 4x2 + 18x - 10
0
= 2x2 + 9x - 5
So I need two numbers that multiply to -10 and add up
to 9…how about 10 and -1!
2x2 + 10x - x - 5 = 0
2x(x + 5) - 1(x + 5) = 0
(2x - 1)(x + 5) = 0
x = ½ or x = -5
x = -5 isn’t possible so the width of the border needs
to be ½ foot.
Alg II - Chapter 5 Word Problems
Pg 262 #100:
The manager of the Buy More is considering repricing a new model of digital camera. At the current
price of $680, the store sells about 70 cameras per
month. Sales data from other stores indicates that for
each $20 decrease in price, an additional 5 cameras
per month would be sold.
A) How much should the manager charge for a
camera to maximize monthly revenue?
Solution:
Revenue = # cameras sold • price
Revenue = (70 + 5x) • (680 - 20x)
Revenue = 47600 - 100x2 + 2000x
Revenue = -100x2 + 2000x + 47600
Max always occurs at the vertex.
x = -b/2a = -2000/-200 = 10
Price = 680 - 20x but now we know x = 10
Price = 680 - 20 • 10 = 680 - 200 = $480
The manager should charge $480
Alg II - Chapter 5 Word Problems
B) What is the maximum revenue?
This is the y-value of the vertex.
(70 + 5(10))(680 - 20(10))
(70 + 50)(680 - 200)
(120)(480) = $57600
The revenue would be $57,600.
Pg 297 #79:
For the years 1989-1996, the amount A (in billions of
dollars) spent on long distance telephone calls in the
U.S. can be modeled by A = 0.560t2 + 0.488t + 51
where t is the number of years since 1989. In what
year did the amount spent reach 60 billion?
Solution: Notice that they’re giving you A and NOT t.
So I need to set the equation equal to 60 and solve.
60 = 0.560t2 + 0.488t + 51
0 = 0.560t2 + 0.488t - 9
Quadratic formula time!
t = -4.46 and t = 3.6.
Negative years doesn’t make sense so we use 3.6.
But remember that t represents the number of years
since 1989. So what year is it then?
1989 + 3.6 = mid-1992