MPM2D Grade 9
QUADRATIC RELATIONS – WORD PROBLEMS
Example 1 A person made a golf shot that went over a tree and into the
cup. The tree was 15 m high and the distance from the green was 150 m.
a) Determine an equation that traces the path of the ball.
Solution
Since the x-intercepts are 0 and d, the equation of the path is
h = a ∙ d (150 – d)
To find a, substitute d = 75, h = 15:
15 = a ∙ 75(150 –75) ⟹ a =
𝟏
𝟑𝟕𝟓
.
Therefore,
h=
𝟏
𝟑𝟕𝟓
d (150 – d)
b) Determine the height of the ball when it was 30 m from the hole.
d = 150 – 30 = 120,
h=
𝟏
𝟑𝟕𝟓
(120)( 150 –120) = 9.6 m.
Example 2 Max operates a store that sells T-shirts. In order to increase
revenue, he decides to increase his price. He knows that over the last six
months he has sold an average of 30 shirts per week at $10 each. Market
research indicates that for every $1 increase in price, weekly sales will
drop by 1 shirt. What unit price will maximize Max’s daily profit?
a) What variable should be included in an algebraic model of the
situation? How will the variables be related?
Answer:
$x – increase in price,
$p – price of T-shirts, p = 10 + x,
n – weekly sales of T-shirts, n = 30 – x.
b) Write an algebraic expression to model the total weekly revenue in
terms of one variable.
Answer:
R = (10 + x)(30 – x)
c) Graph the relation.
Answer:
Increase in
price, x ($)
Price per T-shirt
p = 10 + x ($)
0
5
10
15
20
25
30
10
15
20
25
30
35
40
Number of Tshirts sold,
n = 30 – x
30
25
20
15
10
5
0
Revenue, R ($)
300
375
400
375
300
175
0
d) What unit price will maximize Max’s daily profit?
Answer:
Rmax = $400 when x = $10, p = $20.
e) Why are there two points on the graph that correspond to weekly
revenue of $0?
Answer:
The parabola which branches are directed down and the vertex is above
the x-axis has two zeros.
f) Why does only one of these zeros make sense in the context of the
problem?
Answer:
Only one of these zeros makes sense in the context of the problem,
because the other zero is negative.
APPLICATION PRACTICE
1. The second span of the Bluewater Bridge, in Sarnia, Ontario, is
supported by a pair of steel parabolic arches. The arches are set in
concrete foundations that are on opposite sides of the St. Clair River 281
m apart. The top of each arch rises 71 m above the foundations.
Determine an algebraic expression that models the arch.
Solution
Since the x-intercepts are –140.5 and d, the equation of the path is
h = a ∙ (d –140.5) (d +140.5)
To find a, substitute d = 0, h = 71:
71 = a ∙ (0 –140.5) (0 +140.5) ⟹ a = – 0.0036
Therefore, h = – 0.0036 (d –140.5) (d +140.5).
2. Neil sets the prices at the hardware To Go store. His research shows
that an increase of $0.10 in price of a package of batteries causes a drop
in sales of 10 packages per day. The stores normally sell 600 packages
of batteries per day, at $5.00 per package.
a) What is the maximum revenue Neil can expect on the sale of
batteries?
Solution
Let the increase in price of a pachage be $0.10n, then the new package
price would be $(5.00 + 0.10n), the new sales per day would be (600 –
10n) packages, and the new daily revenue would be
R = (5.00 + 0.10n) (600 – 10n)
dollars. Determine the maximum revenew.
R = 0 ⟹ 5.00 + 0.10n = 0 or 600 – 10n = 0
n = – 50 or n = 60
The coordinates of the vertex:
− 𝟓𝟎 + 𝟔𝟎
n=
= 5,
𝟐
R = (5.00 + 0.10 ∙ 5) (600 – 10 ∙ 5) = (5.50) (550) = 3 025.
The maximum revenew is Rmax = $ 3 025.
b) How many packages of batteries must be sold to generate the
maximum revenue?
Solution
600 – 10n = 600 – 10(5) = 550.
Thus, 550 packages of batteries must be sold to generate the maximum
revenue.
c) What price will generate the maximum revenue?
Solution
5.00 + 0.10n = 5.00 + 0.10 ∙ 5 = 5.50.
Thus, the price that will generate the maximum revenue is $5.50.
3. A toy rocket is launched from a rooftop. It reaches a maximum height
of 125 m at 3 seconds and lands on the ground at 8 seconds.
a) What are the zeros of this relation?
Solution
One of the zeros is 8 s. Determine the other zero. Since the t-coordinate
of the vertex is 3, we get:
𝒕+ 𝟖
= 3 ⟹ t + 8 = 6, t = – 2.
The zeros of this relation are (–2, 0) and (8, 0).
𝟐
b) Determine an algebraic expression to model this relation.
Solution
Since the t-intercepts are –2 and 8, the equation of the relation is
h = a (t + 2) (t – 8)
To find a, substitute t = 3, h = 125:
125 = a (3 + 2)(3 – 8) ⟹ a = –5.
Therefore,
h = –5 (t + 2) (t – 8)
= –5 (t2 + 2t – 8t – 16)
= –5t2 + 30t + 80.
4. Jason is building a rectangular garden. He has 24 m of wire fencing
available.If he must fence around all four sides of the garden, what is the
maximum area his garden can be?
Solution
Let l be the length and w be the width of the
rectangular garden. Since the perimeter of the
garden is 24 m, we get:
P = 2(l + w) ⟹ 24 = 2(l + w)
12 = l + w
w = 12 – l
The area of the garden is A = lw = l(12 – l). Since l > 0 and w > 0, we
get: 0 < l < 12.
Determine the maximum area of the garden.
A = 0 ⟹ l(12 – l) = 0 ⟹ l = 0 or l = 12.
The coordinates of the vertex:
𝟎 + 𝟏𝟐
l=
= 6,
𝟐
A = l(12 – l) = (6)(12 – 6) = 36.
The maximum area the garden is 36 m2.
6. Jason decides to have the garden against the side of his house so that
he only has to fence three sides. What are the dimensions and the area of
the largest garden he can now have?
Solution
Let l be the length and w be the width
of the rectangular garden. Since the
perimeter of the garden is 24 m, we get:
P = l + 2w ⟹ 24 = l + 2w
l = 24 – 2w
The area of the garden is
A = lw = w(24 – 2w).
Since l > 0 and w > 0, we get: 0 < w < 12.
Determine the maximum area the garden.
A = 0 ⟹ w(24 – 2w) = 0 ⟹ w = 0 or w = 12.
The coordinates of the vertex:
𝟎 + 𝟏𝟐
w=
= 6 m, A = w(24 – 2w) = (6)(24 – 2 ∙ 6) = 72 m2.
𝟐
The maximum area the garden is 72 m2.
The dimensioms of the garden are:
w = 6 m, l = 24 – 2w = 24 – 2 ∙ 6 = 12 m.
6. Elaine must set the ticket price for a comic book convention. She
knows from past events that a decrease of $1.50 will result in 75 more
tickets sold. Last year, the convention ticket cost $37.50 and 825 people
attended.
a. What ticket price will result in zero revenue?
Solution
Let the increase in price of a ticket be $1.50n, then the new ticket price
would be $ (37.50 – 1.50n), the new number of the tickets sold would be
(825 + 75n), and the new revenue would be
R = (37.50 – 1.50n) (825 + 75n)
dollars. Determine the ticket prices that will result in zero revenue:
R = 0 ⟹ 37.50 – 1.50n = 0 or 825 + 75n = 0
n = 25 or n = –11
If n = 25 then 37.50 – 1.50n = 37.50 – 1.50(25) = 0.
If n = –11 then 37.50 – 1.50n = 37.50 – 1.50(–11 )
= 37.50 + 16.50
= 54.
The ticket price that will result in zero revenue are $0 and $54.
b. What ticket prices will maximize the revenue?
Solution
The coordinates of the vertex:
− 𝟏𝟏 + 𝟐𝟓
n=
= 7,
𝟐
R = (37.50 – 1.50 ∙ 5) (825 + 75 ∙ 5)
= (30.00) (1 200) = 36 000.
The maximum revenew is Rmax = $ 38 000.
The ticket prices that will maximize the revenue is
37.50 – 1.50(7) = $ 27.00
Answers
1. h = – 0.0036 (d –140.5) (d +140.5).
2. a) $3 025.00
b) 550
c) $5.5
3. a) (–2. 0) and (8, 0), b) h = –5t2 + 30t + 80
4. 36 m2
5. 12 m × 6 m, 72 m2
6. a) $0 and $54.00
b) $27.00