G3 Using Graphs to Solve Activities Equations

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G3 Using Graphs to Solve Activities Equations
Mathematics SKE, Strand G
UNIT G3 Using Graphs to Solve Equations: Activities
G3 Using Graphs to Solve
Equations
Activities
Activities
G3.1
Maximum Volume of an Open-Topped Box
G3.2
Dipstick Problem
Notes and Solutions (2 pages)
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UNIT G3 Using Graphs to Solve Equations: Activities
Mathematics SKE, Strand G
ACTIVITY G3.1
Maximum Volume of an Open-Topped Box
We want to make cardboard trays for
displaying home-made goods at a local
bazaar.
The sheets of card available are of size
10 cm × 12 cm
10 cm
If we cut out a one cm 2 square from each
corner and fold the sides up, we have a tray
of volume
8 × 10 × 1 = 80 cm 3
1 cm
Can you form a tray with a larger volume?
12 cm
1.
Construct trays by cutting out squares of side 2 cm, 3 cm, etc.
For each tray, determine the volume. Which tray has maximum volume?
x
We can find an even larger volume by
considering cutting a square of side x cm
from the card, as shown opposite.
2.
x
What is the length and breadth of the base
of the tray?
Show that the volume is given by
V = 4 x 3 − 44 x 2 + 120 x
3.
Draw a graph of V against x, plotting V for
x = 0. 0.5, 1, 1.5, ..., 5
x
x
Estimate the value of x which gives the
maximum value.
We can generalise these results further by considering the same problem of finding the tray
of maximum volume that can be cut from a square of side a cm.
4.
a
Show that the volume of the tray is
given by
V = 4 x 3 − 4a x 2 + a 2 x
x
We can non-dimensionalise the problem
by taking a = 1 .
a
x
x
x
5.
With a = 1 , plot a graph of V against x
for
x = 0, 0.1, 0.2, ..., 0.5
What value of x gives a maximum value
for V ?
© CIMT, Plymouth University
UNIT G3 Using Graphs to Solve Equations: Activities
Mathematics SKE, Strand G
ACTIVITY G3.2
Dipstick Problem
Service stations very rarely run out of
fuel.
This is due partly to efficient deliveries
but also to precise stock control.
Each type of fuel (petrol and diesel) is
stored in an underground tank and the
amount in each tank is carefully
monitored using some form of dipstick.
It is easy to measure the height, say h, of fuel in the tank. However, the volume will be
proportional to the cross-sectional area – not the height. Suppose the cross-section is a
circle (it is, in fact, usually elliptical, but a circle is a good approximation). You need to
find the relationship between area, A, and height, h, and so provide a ready reckoner to
convert height to area.
For simplicity, take r = 1 m . For values of h
from 0 to 1, you need to find the angle θ and the
area of fuel.
O
θ
r
Problem 1
Show that cos θ = 1 − h .
Problem 2
Show that the area of the sector
OAB is given by
πθ
180
Problem 3
Show that the area of the triangle
OAB is
r−h
B
A
h
(1 − h) sin θ
Problem 4
Deduce the area of the cross-section of fuel and express this as a fraction, A′ ,
of the complete cross-sectional area of the tank.
Problem 5
(a)
(b)
Problem 6
Problem 7
Using the equation in Problem 1,
find the value of θ for each value
of h in the table opposite.
Use the formula deduced in
Problem 4 to find the area
fractions.
h
θ°
0
0
Area fraction
0
0.1
25.84
0.019
0.2
...
...
Plot a graph of A′ (vertical axis)
against height h (horizontal axis).
...
...
...
...
...
...
Use your graph to estimate the height
that corresponds to an area fraction of
1.0
90.00
0.500
(a)
0.05
(b)
0.10
Extension
Construct a dipstick for this problem, from which you can read off the area fraction against
every height value.
© CIMT, Plymouth University
UNIT G3 Using Graphs to Solve Equations: Activities
Mathematics SKE, Strand G
ACTIVITIES
G3.1 and G3.2
Notes and Solutions
Notes and solutions given only where appropriate.
G3.1 One of the best ways to approach this problem is to set it as a practical exercise for group
work.
1.
x
1
2
V 80
2.
3
96
72
4
5
32
0
x = 2 gives maximum volume.
l = 12 − 2 x, b = 10 − 2 x and
V = x (12 − 2 x ) (10 − 2 x )
(
= x 120 − 44 x + 4 x 2
)
= 120 x − 44 x 2 + 4 x 3
3.
4.
The graph should give a value of x about 1.8.
V = x (a − 2 x ) (a − 2 x )
(
= x a 2 − 4a x + 4 x 2
)
= a 2 x − 4a x 2 + 4 x 3
5.
With a = 1, V = x − 4 x 2 + 4 x 3 and x ≈ 0.17 gives the maximum value of V.
G3.2 1. From the triangle, cos θ =
1− h
=1− h
1
2θ ⎞
θπ
2. Cross-sectional area = ⎛
× π12 =
⎝ 360 ⎠
180
1
3. Area of triangle = 2 × ⎛ × 1 × (1 − h) × sin θ ⎞ = (1 − h) sin θ
⎝2
⎠
4. Area of fuel cross-section =
Area fraction =
© CIMT, Plymouth University
θπ
− (1 − h) sin θ
360
θ
(1 − h) sin θ
−
360
π
Mathematics SKE, Strand G
ACTIVITIES
5.
UNIT G3 Using Graphs to Solve Equations: Activities
G3.1 and G3.2
h
θ°
Area fraction
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0
25.84
36.87
45.57
53.13
60.00
66.42
72.54
78.46
84.26
90.00
0
0.019
0.052
0.094
0.142
0.196
0.252
0.312
0.374
0.436
0.500
Notes and Solutions
6.
0.5
0.4
A′ 0.3
0.2
0.1
0
0.2
0.4
0.6
0.8
1.0
h
7. (a) Approximately 0.2
(b) Approximately 0.32
Extension
This is the lower half of the dipstick
Height
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Area

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