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Optimization of functions




Item 4.2 Extended Modeling & Problem-Solving Task




Queensland Government


August 16, 2013

Authored by: KULESHOVA, Anastasiia




Question 1

  1. functions where a is a rational number


Figure 1
For functions f(x)=x2 and f(x)=x4 gradient functions will be f ’(x)=2x and f ‘(x)=4x3 , and at f(x)=0 both functions have only one answer: x=0. This means that other functions with even numbers will have a minimum at x=0. Functions where a is even have a parabola shape with minimum value at x=0 (see Figure 1 graph)


f(x)=x100

f(x)=x10

f(x)=x4

f(x)=x2
http://yotx.ru/graph.ashx?clr0=000000&exp0=x%5e2&clr1=000099&exp1=x%5e4&clr2=ff0000&exp2=x%5e10&clr3=990000&exp3=x%5e100&mix=-2&max=2&asx=on&u=mm&nx=x&miy=0&may=2&asy=on&ny=f%28x%29&iw=600&ih=400&ict=png&aa=on


Figure 2

I have chosen function f(x)=x10 as general shape function because all positive rational numbers before 10 have smooth graphs, and all numbers after 10 have similar shape with almost sharp angles of the graph.


General shape of the function with a positive rational number instead of a and minimum at x=0 will be parabola. To build a general shape I compared few functions and came to conclusion that functions f(x)=x100 and bigger have almost the same shape, so, general shape for the parabola function: Figure 2


f(x)=x10
http://yotx.ru/graph.ashx?clr0=000000&exp0=x%5e10&mix=-2&max=2&asx=on&u=mm&nx=x&miy=0&may=2&asy=on&ny=f%28x%29&iw=600&ih=400&ict=png&aa=on



Figure 3
For functions f(x)=x3 and f(x)=x5 gradient functions will be f ’(x)=3x2 and f ’(x)=5x4 , and at f(x)=0 both functions have only one answer: x=0. To find what stationary point it is we need to find second derivative of this functions: f ’’(x)=6x and f ’’(x)=20x3 , f ’’(0)=0 for both functions – so this stationary point is a point of inflection. Similarity of those two functions means that other functions with odd numbers will have a point of inflection too. Functions where a is odd have a cubic parabola shape with minimum value at x=0 (see Figure 3)


f(x)=x101

f(x)=x11

f(x)=x5

f(x)=x3
http://yotx.ru/graph.ashx?clr0=000000&exp0=x%5e3&clr1=330099&exp1=x%5e5&clr2=ff0000&exp2=x%5e11&clr3=990000&exp3=x%5e101&mix=-1.5&max=1.5&asx=on&u=mm&nx=x&miy=-1.5&may=1.5&asy=on&ny=f%28x%29&iw=600&ih=400&ict=png&aa=on


Figure 4

I have chosen function f(x)=x11 as general shape function because all positive rational numbers before 11 have smooth graphs, and all numbers after 11 have similar shape with almost sharp angles of the graph.


General shape of the function with a positive rational number instead of a and point of inflection will be cubic function. To build a general shape I compared few functions and came to conclusion that functions f(x)=x101 and bigger have almost the same shape, so, general shape for the parabola function: Figure 4


f(x)=x11
http://yotx.ru/graph.ashx?clr0=000000&exp0=x%5e11&mix=-1.5&max=1.5&asx=on&u=mm&nx=x&miy=-1.5&may=1.5&asy=on&ny=f%28x%29&iw=600&ih=400&ict=png&aa=on


Figure 6

Figure 5
Functions f(x)= with positive fractional values instead of a have two general shapes:


f(x)=



f(x)=




Figure 5 outlines shape for fractional values with even denominator and Figure 6 outlines shape for fractional values with odd denominator.

On the assumption of two general shapes for fractional values instead of a it is reasonable to state that all fractional values with odd denominator will be defined for all x (see figure 6).

Referring to general shapes for fractional functions will help to investigate gradient at x=0:

For f(x)=

f ’(x)= and at x=0 f ’(0)= undefined, so there is no gradient there

This will happen with all even dominators at fractional values, the shape of the graph for these values confirms this assumption.

For f(x)=

G-calc for f ’(x)= when x=0



Value of the gradient increases till 933.892 – this is a gradient for f ’(x)=, after this number gradient stays the same, before this number gradient increases from 100.

Two general shapes for function with fractional values instead of a show that none of these functions have minimum at all, so it is impossible to have minimum at x=0 for functions with fractional values with numerator=1 instead of a.



Figure 8

Figure 7
Functions f(x)= with positive fractional values instead of a: for example a= , these fractions have two general shapes (see Figure 7 and Figure 8)


f(x)=



f(x)=



f(x)=



f(x)=




Figure 7 and Figure 8 show consistent pattern that fractions where numerator is even and denominator is odd will have a graph which doubles itself after crossing y-axis. Fractions where numerator is odd and denominator is even will have a graph with only x-positive values.

f(x)=

f(x)=

Index for the radical fraction is odd and this means that domain for this function is [∞;∞]

f(x)=

f(x)=

after that

This means that the domain of this function is [0;∞]

x

f(x)=

f(x)=

-2

1.6

undefined

-1

1

undefined

0

0

0

1

1

1

2

1.6

1.7

Calculus, table and graph evidence confirm that functions f(x)= and f(x)= have different domains.



  1. There are other possible f(x)= with fractional values instead of a:

If values will be negative – graphs of the function will become hyperbolic


Figure 9
If fraction will have even numerator and odd denominator which is less then numerator function will have parabolic shape and will have minimum point: see Figure 9


f(x)=

Derivative function will be: f ’(x)=

When f ’(x)=0 x=0

When x=0 f(0)= f(0)=0

Stationary point for f(x)= function is (0;0)



f(x)=






Figure 10
If fraction will have odd numerator and odd denominator function will have a cubic function shape and point of inflection: see Figure 10


f(x)=

Derivative function will be: f ’(x)=

When f ’(x)=0 x=0

When x=0 f(0)= f(0)=0

Stationary point for f(x)= function is (0;0)



f(x)=




Conclusion:

  • Any Stationary points for function f(x)= are at (0;0) point or there is no Stationary Point at all.

  • To make a function with a fraction instead of a and with [∞;∞] domain – denominator should be odd number.

  • Fractions instead of a with numerator 1 have no minimum.



Question 2

  1. functions


Figure 12

Figure 11

f(x)=



f(x)=


If to take a simple example of the y= function as y= where one positive value is even and its separate curve will be parabola shape, and another positive value is odd which has cubic function shape (Figure 11), composition of this values will change their curves (Figure 12).


f(x)=


y= function is a combination of quadratic and cubic functions. This function has local minimum and maximum its shape is a cubic function with point of inflection but which has not a gradient zero.http://yotx.ru/graph.ashx?clr0=6600ff&exp0=x%5e2+%2b+x%5e3&mix=-4&max=4&asx=on&u=mm&nx=x&miy=-5&may=5&asy=on&ny=y&iw=600&ih=400&ict=png&aa=onhttp://yotx.ru/graph.ashx?clr1=ff0000&exp1=x%5e2&clr2=000000&exp2=x%5e3&mix=-4&max=4&asx=on&u=mm&nx=x&miy=-5&may=5&asy=on&ny=y&iw=600&ih=400&ict=png&aa=on


f(x)=



f(x)=



Figure 14

Figure 13
In the example when function y= has both numbers even shape of such function will depend on negative and positive values, but we all know that a separate curve of the even number instead of a will be parabola. For example function y= will have different shapes depending on negative and positive signs in the fraction (see figure 13 and 14)http://yotx.ru/graph.ashx?clr0=000099&exp0=x%5e2+-+x%5e4&clr1=ff0000&exp1=-x%5e2+-+x%5e4&mix=-2&max=2&asx=on&u=mm&nx=x&miy=-5&may=1&asy=on&ny=y&iw=600&ih=400&ict=png&aa=on


f(x)=



f(x)=



f(x)=



f(x)=


In the case when both values are positive or both negative their curve is an ordinary parabola, but when one value is negative and one positive their curve has three stationary points, in case with f(x)=x2-x4 the curve has one maxima and two minima, and when f(x)=-x2+x4 the curve has one minima and two maxima. http://yotx.ru/graph.ashx?clr0=6600ff&exp0=x%5e2+%2b+x%5e4&clr1=ff00ff&exp1=-x%5e2+%2b+x%5e4&mix=-2&max=2&asx=on&u=mm&nx=x&miy=-1&may=5&asy=on&ny=y&iw=600&ih=400&ict=png&aa=on


Figure 16

Figure 15
If to put two odd values into the y= this values will have two possible curves which will depend on the positive and negative numbers as in the example with even numbers. So if to take the y= function and change cases with positive and negative signs – shapes of the curve will change as well:


f(x)=


http://yotx.ru/graph.ashx?clr0=d92525&exp0=-x%5e3+-+x%5e5&clr1=000000&exp1=-x%5e3%2bx%5e5&mix=-1.5&max=1.5&asx=on&u=mm&nx=x&miy=-1&may=1&asy=on&ny=y&iw=600&ih=400&ict=png&aa=on
f(x)=


http://yotx.ru/graph.ashx?clr0=d92525&exp0=x%5e3+%2b+x%5e5&clr1=000000&exp1=x%5e3-x%5e5&mix=-1.5&max=1.5&asx=on&u=mm&nx=x&miy=-1&may=1&asy=on&ny=y&iw=600&ih=400&ict=png&aa=on


f(x)=






f(x)=





Figures 15 and 16 show that in the case when both values are positive or both negative their curve is an ordinary shape of a cubic function, but when one value is negative and one positive their curve has three stationary points, in case with f(x)=x3-x5 the curve has one maxima, one minima and one point of inflection, and f(x)=-x3+x5 function has the same amount of stationary points but they are reflected along y-axis.


Figure 17
There is another pattern with values in the y= function: if values become bigger the curve becomes more sharp and if and values are far different (as 3 and 55) the curve’s gradient will be broader and if values are close to each other (as 51 and 57) the curve’s gradient will be slim (see figure 17)


f(x)=



f(x)=



f(x)=


http://yotx.ru/graph.ashx?clr0=000000&exp0=x%5e5+-+x%5e3&clr1=330099&exp1=x%5e55+-+x%5e3&clr2=ff0000&exp2=x%5e57+-+x%5e51&mix=-1.1&max=1.1&asx=on&u=mm&nx=x&miy=-0.8&may=0.8&asy=on&ny=y&iw=600&ih=400&ict=png&aa=on



  1. Curves y= can have cusp the fraction instead of a has a fraction with even numerator and odd denominator


Figure 18
Curve with a shift along x and y-axis: (figure 18)


f(x)=


f(x)= function has a cusp due to even numerator over odd denominator, (x-4) in brackets represents a shift to the right along x-axis, +2 at the end represent a shift up along the y-axis



Figure 19
Inverse curve with shift along x and y-axis: (figure 19)


f(x)=


Negative sign in front of the f(x)= equation represents the inverse of the graph upside down, (x+7) in brackets represents a shift to the left along x-axis, -5 at the end represent a shift down along the y-axis, a fraction made a curves range very small that is why it looks almost as a straight line because 2 and 23 are far from each other, it looks like that the cusp is at (-7; -5.3) point, but table of values shows that the cusp as at (-7;-5), so it is reasonable to assume that slim graph is distorted.


Figure 20
Curves with cusp and straight line: (figure20)


f(x)=


Negative sign in front of the f(x)= equation represents the inverse on the graph upside down, (x-3) in brackets represents a shift to the right along x-axis, -3 at the end represent a shift down along the y-axis, a fraction change a curve from smooth to sharp straight line due to big values which are close to each other


f(x)=



Figure 21
An ordinary function f(x)= which has a cusp at x=0 (Figure 21) can be modified by adding or subtracting values.

^ Features of the f(x)= function: (Figure 22)


f(x)= + x



f(x)=



Figure 22
By adding x to the function with a cusp the shape of the graph changes: the pattern of the new graph is that it adds an x value of f(x)= to its y value

Eg.1 Point (1; 1) of the f(x)= will become:

1+1=2

Point (1; 2) of the f(x)=

Eg.2 Point (-3; 2) of the f(x)= will become:

-3+2=-1

Point (-3;-1) of the f(x)=



Figure 23



f(x)= + 2


^ Features of the f(x)= function: (Figure 23)

Some function + or – some number means that that function will shift up or down.


f(x)=


Function f(x)= +2 shifts up by two values along the y-axis


^ A curve with a cusp at x=0 and x=1:


Figure 24
As it is already investigated that functions with a cusp should have a fraction as a power, a fraction needs to have even numerator and odd denominator. To get a cusp at x=0 the function should have x to the power of the investigated fraction, to get a function with a cusp at x=1 x needs to take 1 to make a shift along x-axis.

Figure 24 has two graphs:


f(x)=


f(x)= which has a cusp at x=0 and its shape is almost straight because 22 and 23 values are big and they are close to each other


f(x)=


f(x)= which has a cusp at x=1 and its shape is smooth because 2 and 3 values are small and it has a broad gradient because values are close.


f(x)=



Figure 25
To get a curve with two cusps at x=0 and x=1 two investigated functions above need to be summed up, other functions with other values except of (x-1) will create required curve as well.


f(x)=



f(x)=


A new curve is created by adding value of one curve to another (Figure 25)

Eg.1 f(x)= has a point (0; 0)

f(x)= has a point (0; 1)

0+1=1

So the point at x=0 of the

f(x)= will be (0; 1)

Eg.2 f(x)= has a point (2; 1) f(x)= has a point (2; 2) 1+2=3

So the point at x=0 of the f(x)= will be (2; 3)



  1. functions with combination of features


Figure 26
To get few combinations in one curve I will summarize two functions: with a cusp and a function with one point of inflection, one maxima and one minima


f(x)=



f(x)=


To get proper curve proper values should be used. To make a curve with one point of inflection, one maxima and one minima with a shift along the x-axis, both and fractions need to have the same shift, otherwise the curve of the subtraction of the functions will not have a proper shape (Figure 26)


Figure 27

f(x)=
f(x)= and f(x)= functions have different stationary point values, they do not match so the curve of their sum has no point of inflection when f(x)= values take f(x)= values


f(x)=


The graph I want will be:


f(x)=



f(x)=
f(x)= a curve with one point of inflection, one maxima and one minima with a shift to the right by two values along the x-axis (Figure 27)

To add a cusp curve to the

f(x)= function the cusp curve should have the same shift along the x-axis because the cusp and the point of inflection should coincide with each other or cusp needs to coincide with any point of the other curve to create a cusp in a new curve

All functions can be modified in different cases, but in the situation when I want to get a cusp, and at least some other clear points I will create a cusp in the middle and use a normal cusp curve f(x)= to clearly outline newly created stationary points.



Figure 28
Two functions: f(x)= and f(x)= will be combined by subtraction


f(x)=
f(x)=


f(x)=


+1 at the end of the equation shifts all curves up by one value (Figure 28)


f(x)=
Newly created graph has four stationary points:

f(x)=






X= 2

X=1.0577

X= 2.2222

X= 2.93149



  • (2)= - 2/3 (local max)

  • (1.0577)= 147.4243 (local min)

  • (2.2222)= 0.6665 (local min)

  • (2.93149)= -88.68889 (local max)

At x=2 the curve has a cusp, and at x=1.0577 (local min), x=2.2222 (local min), x=2.93149 (local max)

(Complex calculations were made at http://www.wolframalpha.com/)


Question 5

Gumnut gallery

This report will investigate the distance viewers should be advised to stand away from a wall on which paintings are hung in order to optimise their viewing angle


Figure 29
To find the maximised angle we need do investigate a formula which will work for all cases and scenarios

Figure 29 has a sketch of few patterns:

x - is a distance between person and painting

h – is a height of the eye level of the person

^ L – as a height placement of the painting on the wall

l – is a difference between h and L

Ph – is a size of the painting

θ – viewers angle on the painting

β – viewers angle on the distance between painting and eye level

αθ + β, angle from eye level to the highest point of the painting

As it is clearly provided on the figure 29 a triangle with α angle has a straight angle, triangle with β angle has a straight angle as well, the wanted angle which should be maximised is θ, but this angle is a part of the triangle which does not have a straight angle. A straights angle is to find a formula for unknown angle through triangles sides – opposite, adjacent and hypotenuse.

Sin = Cos = Tan =

So to find α we will need two sides: size of the painting + is a difference between h and L – opposite side and distance between person and painting – adjacent side which is x and should be found as well. Formula for the α will be:

α = Tan-1 (

α = Tan-1 (

Knowing that α is a sum of two smaller triangles we can find θ triangle trough this pattern. As triangle with β angle has a straight angle as well as α triangle We can work through Tan formula to find β

β = Tan-1 (

β = Tan-1 (

Now it is possible to find θ through α taking β

θ = α – β

θ = Tan-1 ( Tan-1 (

If to denote θ as f(x) there will be an equation which will help to find maximised x to make a maximised θ

f(x) = Tan-1 ( Tan-1 (

To use an application to optimisation problems and finding maxima of the curve to state a maximised distance for finding maximised angle the investigated equation will need to be differentiated

Differentiate formula for Tan-1(x):

Let y= Tan-1(x)

Where

-\frac{\pi}{2} < y < \frac{\pi}{2}

Then let Tan y = x

Using implicit differentiation and solving for dv/dx





When 1+Tan2 y = sec2 y




When x = Tan y





So



Differentiated f(x) = Tan-1 ( Tan-1 ( formula will be

f ‘(x) =

Investigated formula will help to find maximised angle for all cases when h is less then L





Figure 30
Ist Scenario

x - is a distance between person and painting

h – is a height of the eye level of the person – 150 cm and 200 cm

^ L – as a height placement of the painting on the wall – 160 cm

l – is a difference between h and L – 10 cm or 40 cm more

Ph – is a size of the painting - 80

θ – viewer’s angle on the painting

β – viewer’s angle on the distance between painting and eye level

αθ + β, angle from eye level to the highest point of the painting

Figure 30 represents a case when person’s height is 150 cm


Figure 31
By substituting values to the equation we will find a maximised distance for the person with 150 cm height

f(x) = Tan-1 ( Tan-1 (


f(x) = Tan-1 ( Tan-1 (
f(x) = Tan-1 ( Tan-1 (

Figure 31 clearly shows that Maximum point is between x=25 and x=35

f ‘(x) =


f ‘(x) =



Figure 32
when f ‘(x)=0

x= -30 and x=30 (figure 32)

We need only positive number as we need maximised value

So when distance from the picture is 30 cm person will see the picture with maximised angle, we will find this angle through already found x value.

θ = Tan-1 ( Tan-1 (

θ = 0.9272 Radians

θ = 0.9272 = 51.5662゚


Figure 33
For the case when person is 200 cm high the angle and distance should be different because person’s height is bigger than a height placement of the painting on the wall. (figure 33)

In this case α is in front of the size of the painting – 80 cm

But other and very important point is that θ triangle has a straight angle but it is now not a full angle of the view


Figure 34
Height of the person is 200 cm, height placement is 160 cm, so person looks straight to the centre of the picture and this means that opposite side of the θ triangle is 40 cm


α = 2Tan-1 (
Formula for finding θ will be: (Figure 34)

θ = Tan-1 (

Differentiated formula will be

θ’ =

β is a straight triangle as well and its opposite side is 40 cm

it is reasonable to state that α will be 2θ

α = 2 Tan-1 (

α’=


α’ =



Figure 35
at α’ =0 equation has an error http://yotx.ru/graph.ashx?clr0=666666&exp0=1%2f1%2b%2840%2fx%29%5e2&mix=-60&max=60&asx=on&u=mm&nx=x&miy=0&may=60&asy=on&ny=y&iw=600&ih=400&ict=png&aa=on

x

α’ =

-1

0.000083

0

undefined

1

0.000083

This means that the angle will decrease by moving from the painting and the optimised viewing angle is 180゚but this is not possible in real life, so

At x=1 θ will be:

α = 2Tan-1 (

α = 177

At the case when person’s eye level height is 150 optimised distance from the painting is 30 cm and angle is 51.5゚; for the case when person’s eye level is at some point at the painting maximised angle should be 180゚, but it is not impossible so the least distance for the viewer can be 1 cm, at one cm the viewing and is 177゚.


Figure 36
At the painting’s size 80 cm and 160 cm of height placement optimised distance for people with height from 150 cm to 200 cm is from 1 cm to 30 cm, the angle will be from 51.5゚ to 177゚
^

IInd Scenario


x - is a distance between person and painting

h – is a height of the eye level of the person – 150 cm and 200 cm

^ L – as a height placement of the painting on the wall – 130 cm

Ph – is a size of the painting - 110

θ – viewers angle on the painting α + β

β – viewers angle on the distance between height placement of painting and eye level

α –angle from eye level to the highest point of the painting (figure 36)

An equation for the scenario two where eye level is on the painting will be:

θ= α + β

Opposite side for β will be h – L

For the case when person height is 150 opposite side for the β is:

150-130=20 cm


Figure 37
Opposite side for α will be the size of the painting – opposite side for β

110-20=90 cm

α= Tan-1 (

β= Tan-1 (

θ= Tan-1 (

Figure 37 shows that the clearest viewing angle for the painting will be 180゚as in case 2 at the 1st scenario.


Figure 38
For the person’s height 200 cm angle will have to change, but it might not happen due to the pattern when eye level is at the painting (figure 38)

Opposite side for β:

200-130=70 cm

Opposite side for α:

110-70=40 cm

α= Tan-1 ( β= Tan-1 (

x

θ ‘(x) =

-1

6502

0

undefined

1

6502

θ= Tan-1 (

Table for the derived function provides the evidence that maximum angle for the viewing is 180゚


Figure 39
To consider all cases and patterns I will investigate a situation when the height of the placement and eye level are the same:

x - is a distance between person and painting

h – is a height of the eye level of the person – 150 cm

^ L – as a height placement of the painting on the wall – 150 cm


Figure 40

Ph – is a size of the painting - 60

θ – viewers angle on the painting

θ = Tan-1 ()

θ = Tan-1 ()

θ‘ =

The graph is the same as cases when eye level is on the painting, this means that optimised angle for this case is 180゚but it is not possible in real life

At the case when person’s eye level height is 150 optimised distance from the painting is 30 cm and angle is 51.5゚; for the case when person’s eye level is at some point at the painting maximised angle should be 180゚, but it is not impossible so the least distance for the viewer can be 1 cm, at one cm the viewing and is 177゚, for the cases wen person’s look goes straight at the painting optimised angle for it is 180゚ at zero distance.

At the cases when viewing person is lower them a height of the placement for 10 cm and when eye level is looking straight at a painting optimised distance for people with height from 150 cm to 200 cm is from 1 cm to 30 cm, the angle will be from 51.5゚ to 177゚

Investigated data gives evidence that mathematical optimisation method is not reasonable for real-world situation due to impossible results as zero distance. For the museum keepers it is not advisable to let people approach the painting closer than 30 cm, but this distance is just the most optimised for a case when the painting is 80cm and it is 10 higher than eye level. The height of the placement should not be high due to standards, but 152 cm will not be proper for optimisation because average people’s height is 160 and optimised angle will be 180゚ which is not reasonable.

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