Mathematics SKE, Strand F
UNIT F6 Solving Inequalities: Activities
F6 Solving Inequalities
Activities
Activities
F6.1
Archimedes' Inequality for π
F6.2
Isoperimetric Inequalities
F6.3
Inequalities for Mean Values
F6.4
Linear Programming
Notes and Solutions (3 pages)
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UNIT F6 Solving Inequalities: Activities
Mathematics SKE, Strand F
ACTIVITY F6.1
Archimedes' Inequality for π
For over 200 years, mathematicians have been attempting to find accurate values for π and
whether this number has any pattern to its decimal form. An early approxiation was given
by Archimedes (in about 250BC) who used 96-sided regular polygons to show that
3
10
1
<π <3
71
7
Here we will follow a quick method.
1.
First start with a square.
(a)
(b)
1
1
90˚
For a circle of radius 1, what is
the area of the inscribed square?
2.
What is the area of the
circumscribed square?
Deduce the upper and lower bounds for π based on your answers to Question 1.
[Hint: the area of the circle is π × 12 = π . )
3.
Now consider a regular 5-sided polygon.
B
P
1
(a)
What is the value of θ ?
(b)
What are the height, OP, and the base length, AB,
of the triangle?
(c)
What is the total area of the inscribed triangle?
(d)
Repeat (a) to (c) but with a circumscribed regular
5-sided polygon.
(e)
Deduce improved upper and lower bounds for π .
A
θ
1
O
Extension
Generalise your method to a regular n-sided polygon.
α
1.
For the inscribed polygon, what is the angle α ?
2.
Find the total area of the inscribed polygon.
3.
What is the area of:
4.
(a)
Deduce bounds for π when:
(b)
In (i) how accurate is your estimate of the value of π ?
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(a)
(b)
O
the inscribed polygon
the circumscribed polygon?
(i) n = 96
(ii) n = 1000
(iii) n = 1000 000
Mathematics SKE, Strand F
UNIT F6 Solving Inequalities: Activities
ACTIVITY F6.2
Isoperimetric Inequalities
According to legend, Princess Dido, a clever Greek princess fleeing from the tyranny of her
brother, landed at Constantinople in Turkey and gained a concession from the local people,
who said that she could have all the land which could be encompassed by the skin of an ox.
She took the biggest ox skin she could find, cut it into very thin, long strips which she
joined together and then placed on the ground, claiming all the land within the skin.
The shape she used to make the enclosed area as large as possible was a circle.
Area,
A
A circle is symmetrical about any diameter and for a given
perimeter, L, the circle gives the maximum area, A, enclosed.
a
Perimeter, L
For a circle of radius a,
A = π a2
Substituting for a:
L2
L
A = π⎛ ⎞ =
⎝ 2π ⎠
4π
Hence for a circle,
4π A
= 1.
L2
L = 2π a , or a =
and
L
.
2π
2
This result is used to measure how close any plane shape is to a circle and the quantity 4π A
L2
is called the Isoperimetric Quotient (IQ) of the shape.
It is conjectured that IQ ≤ 1 , with equality occuring only for the circle.
Questions
1.
2.
Find the IQ for the following plane shapes and show that they support the conjecture above:
(a)
(c)
(e)
(g)
square, side a
rectangle, sides a and 5a
'3, 4, 5' triangle
'5, 12, 13' triangle
(b)
(d)
(f)
(h)
rectangle, sides a and 2a
equilateral triangle
semicircle
regular pentagon.
(a)
Draw up a table of IQ numbers in ascending order.
(b)
What can you conjecture about IQ numbers?
Extension
1.
For a regular n-sided polygon, find an expression for its area and perimeter.
2.
Use these results to find a formula for the IQ of an n-sided polygon.
3.
Evaluate for n = 10 , n = 100 , n = 1000 .
4.
What conjecture can you make?
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UNIT F6 Solving Inequalities: Activities
Mathematics SKE, Strand F
ACTIVITY F6.3
Inequalities for Mean Values
You are probably familiar with the Arithmetic Mean, A, of a set of positive numbers. It is
defined by:
1
A = ( a1 + a2 + a3 + ... + an ) .
n
Two other types of mean values are:
1
(1)
(2)
the Geometric Mean, G, defined by:
G = ( a1 a2 a3 ... an ) n
the Harmonic Mean, H, defined by:
⎛1
1
1
1⎞
H = n⎜ +
+
+ ... + ⎟
an ⎠
⎝ a1 a2 a3
–1
Questions
1.
(a)
For the five values, 1, 2, 3, 4 and 5, evaluate A, G and H.
(b)
For the five values, 3, 3, 3, 3 and 3, evaluate A, G and H.
[Note: When all the values are equal, then A = G = H (= common value).
This is a property which any mean value must have.]
2.
Our conjecture is that for any set of values, A > G > H , and the equality occurs only
when all the values are equal.
Show that this inequality holds for your own choice of values.
3.
For n = 2 , i.e. two values, say a and b,
A =
a+b
,
2
G=
ab ,
H =
2
1 1
+
a b
(a)
Using the fact that
( a − b )2
≥ 0 , show that a 2 + 2 ab + b 2 ≥ 4 ab .
(b)
Hence deduce that
( a + b )2
≥ 4 ab , and that A ≥ G .
Extension
Use a similar method to prove that G ≥ H .
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Mathematics SKE, Strand F
UNIT F6 Solving Inequalities: Activities
ACTIVITY F6.4
Linear Programming
Linear Programming was developed during the Second World War to solve complicated
optimisation problems.
Sample Problem
To find the best (cheapest) way to organise coaches
for a school trip for 560 people (students and staff)
The coach firm contacted has two types of coaches:
Coach type
Capacity
Cost per hour
No. available
Double decker
60
£10
6
Single decker
40
£40
15
Method
Let x = no. of double deckers and y = number of single deckers.
1.
The firm has only 6 double deckers, so 0 ≤ x ≤ 6 .
Write down a similar constraint for y.
2.
(a)
What is the total number of passengers who can be carried in x double deckers
and y single deckers?
(b)
Write down the appropriate inequality to be satisfied.
y
20
3.
4.
On an appropriate set of axes, similar to those opposite,
illustrate all three inequalities on a graph by first
drawing x = 0 , x = 6 , etc.
The boundaries of these lines define the feasible region ,
in which all the inequalities are satisfied.
Show this region by shading.
15
10
5
5.
The cost, C is given by the formula C = 50 x + 40 y .
Lines given by C = constant are straight, parallel lines.
0
5
10
15 x
(a)
Draw C = 1000 , C = 900 , C = 800 , etc. on your graph.
(b)
At what point will C reach its minimum value inside, or on, the boundary of the
feasible region?
(c)
What is the optimum (best) solution to the problem?
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Mathematics SKE, Strand F
ACTIVITIES
UNIT F6 Solving Inequalities: Activities
F6.1 – F6.2
Notes and Solutions
Notes and solutions given only where appropriate.
F6.1 1.
(a)
2
(b)
2.
2<π <4
3.
(a)
36°
(b)
(d)
3.6327
(e)
Extension
α =
360
,
n
4
cos 36° ≈ 0.8090 , 2 sin 36 ≈ 1.1756°
(c)
2.3777
2.3777 < π < 3.6327
180 ⎞
180 ⎞
180 ⎞
n sin ⎛
cos ⎛
, n tan ⎛
,
⎝ n ⎠
⎝ n ⎠
⎝ n ⎠
180 ⎞
180 ⎞
180 ⎞
n sin ⎛
cos ⎛
< π < n tan ⎛
⎝ n ⎠
⎝ n ⎠
⎝ n ⎠
F6.2 1.
2.
(a)
3.139350203 < π < 3.1427146
(b)
3.141571983 < π < 3.141602989
(c)
3.141592654 < π < 3.141592654 (correct to at least 9 decimal places)
(a)
π
≈ 0.785
4
(b)
(e)
π
≈ 0.524
6
(f)
(h)
π tan 54°
≈ 0.865
5
2π
≈ 0.698
9
(c)
2π 2
≈ 0.747
(π + 2)2
5π
≈ 0.413
38
(g)
(d)
π
3 3
≈ 0.605
2π
≈ 0.419
15
One possible conjecture is that IQ → 1 as the shape becomes closer to a circle.
Extension
For an n-sided polygon, IQ =
π
.
π⎞
⎛
n tan
⎝ n⎠
π
π
π
is small and tan ⎛ ⎞ can be approximated by
,
⎝ n⎠
n
n
giving the result that IQ → 1 as n → ∞ .)
(For large n,
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UNIT F6 Solving Inequalities: Activities
Mathematics SKE, Strand F
ACTIVITY
F6.3 1.
3.
F6.3
Notes and Solutions
A = 3 , G ≈ 2.60517 , H = 2.18978
(a)
( a − b )2
≥ 0
⇒
a 2 − 2 ab + b ≥ 0 2
(b)
⇒
a 2 + b 2 ≥ 2 ab
⇒
a 2 + 2 ab + b 2 ≥ 4 ab
⇒
( a + b )2
⇒
a + b ≥ 2 ab
⇒
⇒
Extension
Note that H =
2
1 1
+
a b
we need to show that
a+b
≥
2
A≥G
≥ 4 ab
ab
2 ab
, so to show that G ≥ H ,
a+b
ab ≥
2 ab
a+b
or
a + b ≥ 2 ab (mutiplying both sides by
or
a+b
≥
2
which we know is true.
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=
A = 3, G = 3 , H = 3
ab ,
a+b
)
ab
Mathematics SKE, Strand F
ACTIVITY
F6.4 1.
2.
3.
UNIT F6 Solving Inequalities: Activities
F6.3
Notes and Solutions
0 ≤ y ≤ 15
60 x + 40 y
(a)
(b)
y
60 x + 40 y ≥ 560
x=6
20
15
y = 15
Feasible
region
C = 1000
10
C = 900
C = 800
5
60 x + 40 y = 560
0
4.
5
C = 500
10
15
x
The lines C = constant will attain minimum value at the point x = 6 , y = 5 ,
when C = 500 .
This can be seen by moving the straight line ( C = constant ) across the feasible
region. At C = 500 , only the point (6, 5) is in the feasible region. This tells us
that we should have
6 double deckers,
5 single deckers.
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