Title: Application of Mathematics
Name
Institution

Question one
Volume of figure one
Volume = area * length
Length = 7500M
Area = area of the outer big rectangle minus the area of the curve and the smaller inner rectangle * Area of the bigger rectangle = height * width
=9.6*8
=76.8 M2 * Area under the curve
We use the equation y=aX2 and solve for a with x and y coordinates as (3.6, 4.2), as this are the coordinates on the peak of the curve. a=0.3240740741 Our equation is: y= 0.3240740741x2 * Area under the curve =07.20.3240740741X2
=0.3240740741X33
= 40.32M2
Area of the inner rectangle= height * width
= 7.2 * 2
=14.4 m2
Total area =76.8-14.4-40.32
=22.08M2
VOLUME = 7500*22.08
=165,600M3
Volume of figure two.
The volume is found by multiplying the total area by height .
Height = 7500 M
Area is obtained by subtracting the area of the semicircle and inner rectangle from outer rectangle
Area of outer rectangle= width * height
=12*12.7
=152.4 M2
Area of the semicircle = R2
= * 3.652
=41.853868 M2
Area of the inner rectangle = height * width
= 5*7.3
=36.5 M2
Total area = 152.4- 41.85868-36.5
=74.046132 M2
Volume= 74.046132 M2 * 7500 M
=555,345.99 M3

Volume of figure three.
This is found by multiplying the height by the total area
Total area
Area of the outer rectangle minus the areas of the two 110 degrees circles .
Area of the rectangle = height * width
= 11 * 6
= 66 metres squared
Area of the circular parts =
110360 * 2 * 2.12* * 2= 16.9331844
Total area = 66- 16.9331844
=49.066815 M2
Volume = 49.066815* 7500
= 368,001.117 M3

B) * Figure one
The cost of constructing tunnel one will be the cost of concrete per cubic meter multiplied by the volume of of the tunnel.
= 120* 165,600
= £19,872,000 * Figure two
Cost = 555,345.99 *120
= £66,641,518.80 * Figure three
Cost = 368,001.117 * 120
= £44,160,134.04

. Choice of the tunnel to construct
I would choose to construct the tunnel illustrated by figure three. This is because it would be cheaper in the long run. Although figure one is a cheaper model to construct than figure three, it does not have the bonus benefits that figure three has. Other factors other than price play a vital role in deciding what project to adopt. The other factors could be the end value of a project or the long term profitability of a project. A two way tunnel allows the simultaneous use of the tunnel by two trains headed for different directions. This has the effect of serving a larger group of people than a train that only has one way. This means that the company will generate more revenue than on tunnels which allow passage of only one train at a time. There is also saving of time as people do not have to wait for one train to pass so that they can use the track. It enables the passage of heavy traffic during the day and night. There is also saving of time as one tunnel could branch and lead to another station using a shortcut. This could reduce journey time between destinations.

Question 2.
Let X be the number of first class carriage. Let Y be the number of the standard class carriage.
Let M be the number of seats in each carriage.
The following equations can be deduced from the information given.
X +Y = 11....................... (i)
47X + 76 Y = 9M........... (ii)
18X + 48y = 2M........... (iii)
From equation (i) we get that X=11 – Y
Substituting this value for X in equations (ii) and (iii) we get:
47(11-Y) + 76Y= 9M
18(11-Y) +48Y = 2M
Opening the brackets we get:
517 -47Y + 76Y= 9M
198-18Y+48Y=2M
Adding the like terms together we get:
517 -29Y=9M...... (iv)
198-18Y=2M....... (v)
Multiplying through equation (v) by 4.5 we get
517 -29Y= 9M
891-135Y=9M
Solving these two equations simultaneously we get Y=3.5, and since all seats are full we round off to the nearest whole number and get Y=4.
Solving for X using equation (i) we get that X=7.
Therefore, a combination of six first class carriages, three standard class carriages, one first class engine and one standard class engine gives a combination that maximises profit for the company.
PS: each train must have a total of eleven carriages which are composed of nine full carriages, one first class engine carriage and one standard class engine carriage.

Proof That This Combination Maximises Profit.
Plugging in the values for X and Y
(6*47*8.20) + (1*18*8.20) + ( 1*48*5) + (3*76*5) = 3840
The company makes a profit of £3840
Using another combination of 3 first class carriages, six standard class carriages, one first class engine and one standard class engine gives
(3*47*8.20) + (1*18*8.20) +( 1*48*5) + (6*76*5) =3823.8
Any other combination will gives values lying between £3840 and £ 3823.80 hence our original combination maximises profit.

b)
I would choose a different arrangement for the train. This is because although the combination gives maximum profit for the company, it locks out most of the middle class and low class people who cannot afford to pay for a first class carriage ticket. Being a train company owned by the government, it is then classified as a property of the state which is to be of benefit to all and sundry. It therefore should not lock out people who cannot afford a first class ticket.
Also, the maximum and the minimum profits using the extreme combination of carriages only gives a difference of £16.20. This shows that the company will not make super profits by prioritizing the first class people and leaving out the standard people. It is therefore the obligation of the company to give equal chances to both first class passengers and second class passengers.
Another point to my argument is that the standard class ticket can be afforded by both people living the high life and those who rank low in the society. Therefore if the first class person misses a space he can always buy a standard class ticket. This is not the case for the standard class person.
The people who ride on trains are mostly poor people who cannot afford vehicles or if they have them, they cannot afford to daily fuel them when they are going to work. As a result these people will opt for train rides. On the other hand first class people can afford to fuel their vehicles to work every day and reserving seats for such people whose chances of them taking train rides are minimal will lead to the train having empty seats. This will be a loss for the company.

Question three Distance travelled in kilometres | 00 | 3,120,000 | 6,240,000 | 9,360,000 | 12,480,000 | 15,600,000 | 18,720,000 | Size of the tyre | 844 | 843 | 842 | 841 | 840 | 839 | 838 |
The size of the tyre after different distances have been covered can be summarised and tabulated as follows:

The distances have been obtained by adding 3,120,000 to the previous distance covered since this is the distance after which a tyre gets worn out by one millimetre. The size of the tyres is obtained by subtracting one millimetre since for every 3,120,000 covered the size of the tyre reduces by one millimetre.
Since the size of the tyre is a linear function of the distance travelled, the linear equation that needs to be solved is:
Y=Ax + B, where X is the distance travelled by the train and Y is the size of the tyre after this distance has been travelled.
Substituting for the different values of X and Y and solving for A and B Using simultaneous equations we get:
843=3,120,000A + B.......... (i)
839=15,600,000A + B.......... (ii)
Solving for A and B using the simultaneous equations above we get that :
A = - 0.0000003205128 and B= 844
Hence plugging in these values in our previous equation we get
Y= -0.0000003205128X + 844
b)
For the tyres to be changed a distance of 18,720,000 kilometres need to be covered. The distance between London and Birmingham is 225 kilometres and a train covers a total of 450 kilometres on a journey to and fro London. Therefore the total number of journeys that need to be made is:
18,720,000kilometres/450 kilometres = 41,600
The number of journeys to be made before the tyres wears out is 41,600

References Kaganova, O. & Mckellar,J., 2006.Managing Government Property Assets: International Experiences. Published by Urban InstPr