2- Correction key

Example of an appropriate solution

The supplier’s new constraint is

2x ≥ y or y ≤ 2x

Polygon of constraints

y

x 0

P

Q

R

S

T

Objective function

Z = 0.90x + 1.40y

Vertices of the polygon of constraints

System of

equations Vertices Value of objective function Profit

480

90

=+=

yx

x

Q (90, 390) 0.90(90) + 1.40(390) = 627

$627

(maximum profit before additional

constraint)

yx

x

==

2

90 S (90, 180) 0.90(90) + 1.40(180) = 333 $333

yx

yx

==+

2

480

T(160, 320) 0.90(160) + 1.40(320) = 592

$592

(maximum profit with additional

constraint)

1

90

480

==+

y

yx R(390,90) 0.90(390)+1.40(90) = 477 $477

90

90

==

y

x P(90,90) 0.90(90) + 1.40(90) = 207 $207

Difference in profit

$627 - $592 = $35

Answer: The profit would decrease by $35.

Example of an appropriate method

x = number of road bikes

y = number of mountain bikes

x ≥ 0

y ≥0

x ≥ 10

y ≥ 45

x + y ≤ 80

y ≥ 3x

Objective Function

Max. Profit = 250x + 175y

10 20 30 40 50 60 70 80

80

70

60

50

40

30

20

10

Road Bikes

Mtn Bikes

(10, 45) (15, 45)

(20, 60)

(10, 70)

Points (x, y) Calculation Profit

1. (10, 45)

250(10) + 175(45) $10 375

2. (15, 45) 250(15) + 175(45) $11 625

3. (20, 60) 250(20) + 175(60) $15 500

4. (10, 70) 250(10) + 175(70) $14 750

Answer The maximum weekly profit is $15 500.

2

Example of an appropriate solution

x: number of hours at first job per month

y: number of hours at second job per month

Constraints before

xyyxyxyyx

≥≤+≥+≥≤≥

603004010

Constraints after

xyyxyxyx

≥≤+≥+≥≥

6030010

(20, 40)

(30, 30)

(10, 40)

(10, 20) (15, 15)

Polygon Before

(10, 50)

(30, 30)

(10, 20) (15, 15)

Polygon After

Maximum Before

Vertices S = 6.3x + 8y ($)

A(10, 40) 383

B(10, 20) 223

C(15, 15) 214.50

D(30, 30) 429

E(20, 40) 446

3

Maximum After

Vertices S = 6.3x + 8y ($)

B(10, 20) 223

C(15, 15) 214.50

D(30, 30) 429

F(10, 50) 463

Difference in maximum salary

$463 − $446 = $17

Answer: Murray’s maximum possible salary increased by $17

Note: Students who use an appropriate method in order to determine the constraints, graph the

original polygon, and find its vertices have shown they have a partial understanding of the

problem.

D

B

C

4

5

6

The region described is 134

22

≤+ yx.

The equation of the ellipse is 136100

22

=+ yx.

or any equivalent equation.

The equation of the spacecraft's trajectory is y = -2x + 30.

or any equivalent equation.

Hyperbola

7

8

9

10

Work : (example)

Co-ordinates of O and measure of r

x2 + y2 − 32x − 10y + 279 = 0

in canonic form

(x − 16)2 + (y − 5)2 = 2

thus O(16, 5) and r = 2

Measure of segment AB

)OB (m + )AO (m = AB m 22

)2( + )2( = 22

2 = 4 = 2 + 2 =

Pythagorean theorem

Co-ordinates of point A

1 = AE m = OE m

x-co-ordinate of A = 16 − 1 = 15

y-co-ordinate of A = 4

In isosceles triangle AOB, which is right-

angled, the axis of symmetry of the parabola

divides segment AB, at the point of

intersection E, into two equal parts.

Point A belongs to the circle with centre O

whose equation is known.

Equation of the parabola

The vertex of the parabola is O(16, 5).

A B

C D x

y

//

O

E

11

y = a(x − h)2 + k

4 = a(15 − 16)2 + 5

a = -1

y = -(x − 16)2 + 5

The co-ordinates of point A are used to

calculate the value of a.

x-co-ordinate of points C and D

0 = -(x − 16)2 + 5

x2 − 32x + 251 = 0

x1 = 16 − 5 and x2 = 16 + 5

Measure of segment CD

( ) ( ) 47.452516516CDm ≈=−−+=

Result : Rounded to the nearest hundredth, the distance between C and D is 4.47 m.

Note.- Also accept 4.48 m.

Work : (example)

Distance of eye from body :

4 units

Distance from vertex to body :

value b of the ellipse

Value a of the ellipse :

4 ÷ 2 = 2 units

And a2 + c2 = b2

22 + 42 = b2

4 + 16 = 20 = b2 ⇒ b = 2 5 ≈ 4.5

Translation of semi-circular feet

x2 + 6x + y2 + 16y + 72 = 0

x2 + 6x + 9 + y2 + 16y + 64 = -72 + 9 + 64

(x + 3)2 + (y + 8)2 = 1

Translation (-3, -8) and radius = 1

or

(3, -8) for the right foot

hx

y12

There are 8 units from foot to neck

Total height : 8 + 4.5 = 12.5

Result : The height h of the robot is 12.5 units.

Example of an appropriate method

Consider the hyperbola centred at the origin (h, k) = (0, 0)

2a = 4 cm = diameter of a circle

a = 2

coordinate (x, y) =

2

4 ,

2

8 = (4, 2)

Equation of the hyperbola

b

a 2

2

2

yx2

− = 1

b4 2

y x

22

−

= 1 (x, y) → (4, 2)

b

4

4

162

−

= 1

b

42

-

= -3 ⇒ b2 = 3

4

1

3

44 =

y x

22

−

Calculation for y when x = 3

4 cm

A

C D

B 6 cm

6 cm

O ? ?

x

y

8 cm

13

1

3

44 =

y x

22

−

1.2961. 1

3

443 2

22

y = y = y

±≈⇒⇒

−

2y ≈ 2.58

Answer The measure of BD is 2.58 cm.

Example of an appropriate solution

Circle

Centre: (13, 10) radius: 4 cm

End points of the diameter

(9, 10) and (17, 10)

Hyperbola

The vertices are (9, 10) and (17, 10) and therefore a = 4.

Half of the total length is 7 cm. So the foci are (6, 10) and (20, 10) and therefore c = 7.

Equation of hyperbola

2

222

222

33

47

b

b

bac

=+=+=

∴ ( ) ( )

133

10

16

13 22

=−−− yx

To find the height let x = 6 and find the y coordinate

14

( ) ( )

( ) ( ) ( )( )( )

( )( )1025.8

100625.68

10161089

10165281617

331610167-33

133

10

16

136

2

2

2

22

22

−=±−=

−=−=−

=−−

=−

−−

yy

y

y

y

y

So

25.1825.810

=+=y

and 75.125.810

=−=y

Answer: The height of the frame is 16.5 cm.

Note: Students who use an appropriate method in order to determine a correct equation of the

hyperbola have shown they have a partial understanding of the problem.

Name : _________________________________

Group : _________________________________

Date : _________________________________

568536 - Mathematics

Question Booklet

Secondary 5 students are organizing a fund-raising activity in order to lower the cost of their grad dance.

During a volleyball tournament, they plan to sell

bottles of water and juice at a stand. The following

constraints are to be respected:

- At least 90 bottles of water and at least 90

bottles of juice must be sold.

- They cannot store more than 480 bottles at

the stand.

y

x 0 30

30

Letting x be the number of bottles of water to be sold and y, the number of bottles of juice to be sold, the

students transformed the constraints into the following inequalities:

480

90

90

≤+≥≥

yx

y

x

The profit on each bottle of water is $0.90 and on each bottle of juice it is $1.40.

To determine their maximum profit, the students drew the graph of the polygon of constraints, shown above.

Before delivery, the supplier imposed a new condition: The students had to order a maximum of twice as many

bottles of juice as bottles of water. This new constraint will lower their maximum profit.

By how much would the profit decrease because of the supplier’s new condition?

Show all your work.

1

Wheeler is a producer of mountain bikes and road bikes. Because of its small size, it can build no more than

80 bikes each week. To meet certain conditions in its workshop, it must build at least 45 mountain bikes, and

at least 10 road bikes weekly. To meet consumer demand, it must manufacture at least 3 times as many

mountain bikes as road bikes.

The following is the system of constraints for Wheeler's weekly bike production:

x = the number of road bikes produced weekly

y = the number of mountain bikes produced weekly

x ≥ 0

y ≥ 0

x ≥ 10

y ≥ 45

x + y ≤ 80

y ≥ 3x

For each road bike and mountain bike produced, Wheeler earns a profit of $250 and $175, respectively.

What is the maximum weekly profit that can be earned?

Show your work.

2

Murray plans a trip to New York in July. In order to save money, he works at two different part-time jobs on

weekends. At the first job, he works a minimum of 10 hours per month and at the second, a maximum of

40 hours per month. Murray must work at least 30 hours per month but no more than 60 hours per month. He

must work at least as many hours at the second job as he does at the first. He makes $6.30 an hour at the first

job and $8 an hour at the second job.

Let x: number of hours per month at first job

y: number of hours per month at second job

The initial constraints for this situation are:

xyyxyxyyx

≥≤+≥+≥≤≥

603004010

Because of a shortage of employees, Murray was later advised that he could increase the number of hours he

worked at the second job.

By how much did Murray’s maximum possible salary increase because of the employee shortage?

Show all your work.

3

Michael wants to take part in an airplane race around two pylons, the tops of which are 2 km apart. To have

the best chance of winning, he estimates that the sum of the distances separating him from the pylons should

always be 3 km.

Express algebraically the ideal route for Michael's airplane.

A)

x2 + y2 = 2.25

C)

125.125.2

22

=− yx

B)

x2 − y2 = 2.25

D)

125.125.2

22

=+ yx

2 km

4

An ellipse has two vertices at (4, -3) and (4, -9), and one of its foci at (0, -6).

Which of the following is the equation of the ellipse?

A)

( ) ( )1

9

6

25

4 22

=−++ yx

C)

( ) ( )1

25

6

9

4 22

=−++ yx

B)

( ) ( )1

9

6

25

4 22

=++− yx

D)

( ) ( )1

25

6

9

4 22

=++− yx

Consider a parabola with vertex (-1, 4) and focus (-4, 4).

Which of the following is the equation of the parabola in standard form?

A)

(y − 4)2 = 12(x + 1)

C)

(y − 4)2 = -12(x + 1)

B)

(x + 1)2 = 12(y − 4)

D)

(x + 1)2 = -12(y − 4)

5

6

In the centre of a large shopping mall there is a pool with a small sailboat floating in it. The boat is held by a

4-metre long cord tied to a ring which is attached to the bottom of the boat. The cord is fixed to two posts, A

and B which are 2 metres apart at the water's surface.

Using a second-degree relation, describe the region inside which the sailboat can move. Consider the mid-

point of segment AB as the origin of the graph.

A B

7

The trajectory of a moving marble is an ellipse with centre at the origin.

The sum of the distances from the marble to F1 and from the marble to F2 is 20 cm. The length of the minor

axis is 12 cm.

What is the equation of the ellipse?

Marble

y

xF

1F

2

8

An American spacecraft completed three circular

orbits around the Earth before continuing on its

journey towards the moon following a trajectory

which is tangent to its previous path.

In the adjacent diagram, the Earth's center is represented by point A.

Point B is the spacecraft's position at the moment it began its journey towards the moon.

What is the equation of the spacecraft's trajectory towards the moon?

B(12, 6)

A(8, 4)

y

x

9

A space probe’s flight path is effected by the proximity of two planets. During flight, the absolute difference

between the distance from the space probe to planet A and the distance of the space probe to Planet B is

constant.

What geometric loci matches this description?

Space Probe

Planet B Planet A

10

A sculpture in the garden of a contemporary art museum consists of a circle and a parabola, as shown below.

The vertex of the parabola coincides with the centre O of the circle which has the equation

x2 + y2 − 32x − 10y + 279 = 0

where the unit of measure is the metre.

The parabola, whose axis of symmetry is vertical, is constructed such that angle AOB is a right angle.

Rounded to the nearest hundredth, what distance is there between bases C and D of the sculpture?

Show your work.

A B

C D x

y

//

O

11

Some Mechanical Engineering students decide to build a miniature robot as a class project. Here is their

design :

•

The middle of the single eye is situated at the focus of the semi-ellipse, centre (0, 0). This focus is

located 4 units from the minor axis of the semi-ellipse.

•

The minor axis of the semi-elliptical head has a length of 4 units.

•

The semi-circular main body of the robot is defined by the equation : x2 + y2 − 25 = 0

•

The robot's semi-circular feet are defined by the equations

x2 + y2 + 6x + 16y + 72 = 0 and x2 + y2 − 6x + 16y + 72 = 0

What is the overall height h of the robot?

Show your work.

hx

y

12

A circle tangent to the two branches of a hyperbola is engraved on Fred's rectangular belt buckle. The buckle is

8 cm long and 4 cm wide.

Furthermore, AB // CD , m AB = 6 cm and m CD = 6 cm.

The centre of the circle is located at the origin of the Cartesian plane below.

What is the length of BD ?

Round your answer to the nearest centimetre.

Show all your work.

4 cm

A

C D

B 6 cm

6 cm

O ? ?

x

y

8 cm

13

A modern picture frame is in the shape of a circle between the two branches of a hyperbola, as shown in the

diagram below.

Height

14 cm

The equation of the circle is (x − 13)2 + (y − 10)2 = 16. The centre of the circle and the centre of the hyperbola

coincide. The vertices of the hyperbola are the endpoints of the horizontal diameter of the circle and the

vertical edges of the picture frame pass through the foci of the hyperbola. The total length of the frame is

14 cm.

What is the height of the frame?

Show all your work.

14