# Maths - Investigate how many people can be carried in each type of vessel - Mathematics Essay Example

The teacher has presented me with a mathematical problem in which I must solve and investigate - **Maths - Investigate how many people can be carried in each type of vessel** introduction. It is as follows:

– The ‘Great Outdoors Boating Company’ has 20 large canoes, 5 small canoes and 10 motorboats for hire.

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– The ‘New Canoe Company’ has 10 large canoes, 15 small canoes and 6 motorboats for hire.

– The ‘Ship-u-Like Company’ has 8 large canoes, 4 small canoes and 2 motorboats for hire.

– The ‘Great Outdoors Boating Company’ can carry a maximum of 225 people.

– The ‘New Canoe Company’ can carry a maximum of 179 people.

– The ‘Ship-u-Like Company’ can carry a maximum of 92 people.

Investigate how many people can be carried in each type of vessel.

When we first consult this problem, the first thing to look at is what the problem tells us and what I can determine from it. It is easy to see that it is a problem which contains three undetermined variables. Therefore, I am quite convinced that the route to solving this certain problem lies in the depths of Algebra. However, in order to solve algebra; we must take the numbers from the realistic world into the abstract world of mathematics. Hence, form equations from the problem. Firstly, I will let large canoes be called “x”, let small canoes be called “y” and let motorboats be called “z”. We can then create a set of three equations with this information:

20x + 5y + 10z = 225

10x + 15y + 6z = 179

8x + 4y +2z = 92

Of all the different ways in which I am familiar with in terms of solving simultaneous equations, there are:

– Elimination

– Substitution

– Trial and Improvement

– Graphical Solution

– Matrices

Of these possible methods, I believe that I will use a combination of both elimination and substitution to solve this problem. This is because I believe that in order to get the best possible result in the fastest possible manner, one must utilise as many techniques in effective combination as available. This technique was taught to me in Year 8 by a mathematics teacher and I plan to make full use of it.

Elimination and Substitution

I will begin this problem by simplifying all three equations, thus make life far easier for myself.

(i) 20x + 5y +10z = 225 => 4x + y +2z = 45 (divided by 5)

(ii) 10x + 15y + 6z = 179 => (cannot be simplified)

(iii) 8x + 4y + 2z = 92 => 4x + 2y + z = 92 (divided by 2)

After doing so, I will then subtract (i) from (iii) since they both have a “4x” factor in it, hence I can eliminate the factor of “x”. This as a result gives me an equation of:

4x + y + 2z = 45

4x + 2y +z = 92

y – z = 1 (I will call this equation (iv))

Now that I have formed one equation with “y” and “z”, I now need to form another on using (i) and (ii). I will multiply (ii) by 2 in order to make the factor of ’10x’ common to both equations and then subtract and eliminate.

20x + 5y + 10z = 225

10x + 15y + 6z = 179 => 20x + 5y + 10z = 225

25y + 2z = 133 (I will call this equation (v))

We now take the newly formed equations (iv) and (v) and we will now eliminate one of the factors from them.

25y + 2z = 133

X2) y – z = 1 => 2y – 2z = 2

27y = 135

y = 5

Now that we know one of the factors, we will now start to utilise the technique of substitution in our problem. We will begin by taking one of the equations that was formed earlier (equation (iv)) and will substitute y = 5 into the equation.

y – z = 1

5 – z = 1

-z = -4

z = 4

Now we have already solved two variables of the equation, since y and z have been solved, we can substitute their values into and of the equations we’ve used before in order to solve ‘x’.

4x + y + 2z = 45

4x + 5 + 8 = 45 => 4x = 32

x = 8

Now, the equation has been solved and we can complete our work by proving these. We will substitute them back into the equations that we began with.

(20 X 8) + (5 X 5) + (10 X 4) = 225

(10 X 8) + (15 X 5) + (6 X 4) = 179

(8 X 8) + (4 X 5) + (2 X 4) = 92

Thankfully, the numbers substitute perfectly back into the equation and as a result, we can conclude that our method has worked and thus:

x = 8

y = 5

z = 4

Matrices

After solving the mathematical problem using one method, it is best to solve it via another method in order to enhance one’s mathematical understanding. The next method that will be used to solve this problem with be via Matrices. This is a mathematical concept that is quite new to me and it was discovered after browsing through a few mathematics websites during my search for knowledge upon the investigation. After reading a tutorial, I eventually realised that the use of matrices is perfect for solving this type of equation. The website whereby I obtained the knowledge of how matrices work is on this page: http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/Summary3.html.

First of all, we must write the problem in matrix form:

[20 5 10] [x] [ 225 ]

[10 15 6 ] [y] = [ 179 ]

[8 4 2 ] [z] [ 92 ]

A X = B

Let the three matrices be named “A,’ ‘X’ and ‘B’ as labelled above. In order to find out x, y and z. We have to firstly rearrange the equation for “X”. Since laws of matrices state that AB is not equal to BA, we must rearrange the matrix by multiplying both sides of the equation by A-1. Therefore:

X = A -1B

In order to work out X, the first thing that needs to be found out is the determinant of A. This can be achieved by taking the top row of Matrix A, getting each number eliminating the corresponding rows in columns that the specific number lies in and leaving the remaining numbers in a matrix of their own. For example, firstly we select the number 20.

[20 5 10]

[10 15 6 ] => 20 [15 6]

[8 4 2 ] [4 2]

We do the same with every number in the top row and we get:

[A] = 20 [15 16] -5 [10 6] +10 [10 15]

[4 2 ] [8 2] [8 4 ]

Since Matrices are in the form:

[ + – + ]

[ – + – ]

[ + – +]

The 20 and 10 are positive while the 5 is made negative. The value of ‘A’ can then be found by cross multiplying the numbers in each matrix and subtracting. Thus:

[A] = 20 (15X2 – 6X4) – 5 (10X2 – 6X8) +10 (10X4 – 15X8)

[A] = -540

We now know that the determinant of A is: 1/-540.

The next step is to find the value of [A]T or transpose by simply rewriting the rows as columns:

[20 10 8]

[5 15 4]

[10 6 2]

After finding the transpose, we need to find the co-factors of the matrix. This is achieved by taking the transpose, getting each number of the matrix – eliminating the corresponding rows and columns and then cross multiplying and the subtracting what remains.

[(15X2-6X4) (2X5-10X4) (5X6-10X15) ]

[(2X10-6X8) (2X20-10X8) (6X20-10X10)]

[(4X10-15X8) (4X20-5X8) (15X20-5X10)]

We must then invert the corresponding signs of the integers using this shape:

[ + – + ]

[ – + – ]

[ + – +]

The final result is:

[6 30 -120]

[25 -40 -20 ]

[-80 -40 250 ]

We now know that A-1 is equal to the above result. Since we know that X = A-1B. We can now substitute the numbers we know back into the equation now that we are aware of what the values of A-1 and B are.

[x] [6 30 -120] [225]

[y] = 1/-540 [-25 -40 20] [179]

[z] [-80 40 250] [92 ]

The two matrices on the right hand side of the equation are then multiplied together:

[225X6+179X30+92X-120 ] [-4320]

[225X25+179X-40+92X-20 ] = [-2700]

[225X-80+179X-40+92X250] [ 2160]

Thus, the final equation is:

[x] [-4320]

[y] = 1/-540 [-2700]

[z] [-2160]

And thus, the answers are revealed:

x= 1/-540 X -4320 = 8

y= 1/-540 X -2700 = 5

z= 1/-540 X -2160 = 4

Since the problem has now been solved using two different methods, it is now safe to conclude and prove that the answers are correct. But despite all of the different problems we’ve solved, we can dig deeper into the algebraic and abstract world of mathematics by solving an even more complex equation. Now that we have managed to solve the problem to a given specific example of this type of problem, we want to solve a problem that will work for any type of simultaneous equations that involves three different variables and three different equations. We will call this new problem that faces us the “General Boating Problem”.

“The General Boating Problem”

In order to solve this new problem, we must presume that we are given no information at all apart from the fact that there are three simultaneous equations and three undetermined variables. The different types of boats will still be labelled as “x”, “y” and “z”. The quantities of these vessels and the number of people however are also undetermined – these will be also labelled with different letters of the alphabet. We will write the three equations as:

(i) Ax + By + Cz = J

(ii) Dx + Ey + Fz = K

(iii) Gx + Hy + Iz = L

In order to solve these equations and find a formula for x, y and z that would work for any boating problem, we will use exactly the same approach as previously – we will try to use elimination. Firstly, we will try to eliminate the “x”, so we will take equations (i) and (ii) in an attempt to do so. Since we need the co-efficient of x to be the same in both equations in order to subtract, we will multiply equation (i) by D and equation (ii) by A.

Ax + By + Cz = J => DAx + DBy + DCz = DJ (i)

Dx + Ey + Fz = K => ADx + AEy + AFz = AK (ii)

(ii) – (i) = (AEy – DBy) + (AFz-DCz) = AK-DJ

Factorise: y (AE-DB) + z (AF-DC) = AK-DJ (iv)

This newly factorised equation is now called equation (iv). However, we are unable to eliminate it on its own, therefore we must create another equation in order to create something to eliminate it with, we will obtain this by eliminating x from equations (i) and (iii):

Ax + By + Cz = J => AGx + BGy + CGz = JG (i)

Gx +Hy + Iz = L => GAx + HAy + IAz = AL (iii)

(iii) – (i) = (HAy – BGy) +(IAz -CGz) = AL – JG

Factorise: y (HA-BG) + z (IA-CG) = AL-JB (v)

This new equation can be called (v) and now that we have a pair of simultaneous equations with y and z, we can now eliminate a factor of y from them. However, before doing so, we need to make the co-efficient of “y” the same in both cases:

y (AE-DB) + z (AF-DC) = AK-DJ

multiply by (HA-BG)

=> y (HA-BG)(AE-DB) + z(AF-DC)(HA-BG) = (AK-DJ)(HA-BG)

y (HA-BG) + z (IA-CG) = (AL-JG)

multiply by (AE-DB)

=> y (AE-DB)(HA-BG) + z (AE-DB)(IA-CG) = (AE-DB)(AL-JG)

We once again subtract the two equations from each other eliminating y and creating:

z(AE-DB)(IA-CG) – z(AF-DC)(HA-BG) = (AE-DB)(AL-JG)-(AK-DJ)(HA-BG)

z = [(AE-DB)(AL-JG) – (AK-DK)(HA-BG)] / [(AE-DB)(IA-GC) -(AF-DC)(HA-BG)]

Now that the formula for ‘z’ has been determined, we now need to find ‘x’ and ‘y’. In order to determine ‘y’, we simply need to use the previous two sets of equations which we had obtained after eliminating ‘x’ (i.e. equations (iv) and (v)). However, this time – instead of eliminating the ‘y’, we will now eliminate the ‘z’ factor.

y (AE-DB) + z (AF-DC) = AK-DJ (iv)

multiply by (IA-CG)

=> y (AE-DB)(IA-CG) + z (AF-DC)(IA-CG) = (AK-DJ)(IA-CG)

y (HA-BG) + z (IA-CG) = AL-JB (v)

multiply by (AF-DC)

=> y (HA-BG)(AF-DC) + z (AF-DC)(IA-CG) = (AF-DC)(AL-JG)

Subtract: (v) – (iv)

=> y (HA-BG)(AF-DC) – y (AE-DB)(IA-CG) = (AF-DC)(AL-JG) – (AK-DJ)(IA-CG)

y = [(AF-DC)(AL-JG) – (AK-DJ)(IA-CG)] / [(HA-BG)(AF-DC) – (AE-DB)(IA-CG)]

Since we now have the formula of ‘y’ and ‘z’ we could simple take any of the three formulas we started with and simply substitute them into the equation and rearrange to produce a formula for ‘x’. However, the problem with that method leads to issues concerning messiness and complication. To achieve and easier and much more simplified version of a formula for ‘x’, we will start from scratch and use the elimination method from the beginning as we did to find the formula for ‘z’.

(i) Ax + By + Cz = J

(ii) Dx + Ey + Fz = K

(iii)Gx + Hy + Iz = L

Eliminating Y:

Multiply (i) by E => AEx + BEy + CEz = JE

Multiply (ii) by B => DBx + EBy + FBz = KB

(ii) – (i) => x (DB-AE) + z(FB-CE) = KB – JE (iv)

Multiply (i) by H => AHx + BHy + CHz = JH

Multiply (iii) by B => GBx + HBy + IBz = LB

(iii) – (i) => x (GB-AH) + z (CH-IB) (v)

Eliminating Z:

Multiply (iv) by (CH-IB)=> x (DB-AE)(CH-IB) + z (FB-CE)(LH-IB) = (KB-JE)(CH-IB)

Multiply (v) by (FB-CE)=> x (GB-AH)(FB-CE) + z (CH-IB)(FB-CE) = (LB-JH)(FB-CE)

(v) – (iv)

=> x(GH-AH)(FB-CE) – x(DB-AE)(CH-IB) = (LB-JH)(FB-CE) – (KB-JE)(IB-CH)

x = [(LB-JH)(FB-CE) – (KB-JE)(IB-CH)] / [(GB-AH)(FB-CE) – (DB-AE)(IB-CH) ]

Now, we have finally succeeded in producing a general formula for each variable for any simultaneous equation that involves three variables and has three equations in it. In order to prove that these equations work, we will return from the abstract world to the real world where the problem first existed. We will substitute the numbers from the original problem back into our new equations for ‘x’, ‘y’ and ‘z’.

Let the following letters represent the following numbers in the problem:

A= 20 B= 5 C= 10 J=225

D= 10 E= 15 F= 16 K=179

G= 8 H= 4 I=2 L=92

We will begin by substituting the numbers into the equation for x:

x = [(LB-JH)(FB-CE) – (KB-JE)(IB-CH)] / [(GB-AH)(FB-CE) -(DB-AE)(IB-CH)]

= [(460-900)(30-150)-(995-3375)(10-40)] / [(40-80)(30-150)-(50-300)(10-40)]

= [(-440 X -120) – (-2480 X -30)] / [(-40 X -120) – (-250 X -30)]

= [52800-74400] / [4800-7500]

= -21600 / -2700

= 8

Then for y:

y = [(AF-DC)(AL-JG) – (AK-DJ)(IA-CG)] / [(HA-BG)(AF-DC) – (AE-DB)(IA-CG)]

= [(120-100)(1840-1800) – (3580-2250)(40-80)] / [(80-40)(120-100) – (300-50)(40-80)]

= [800- (-53200)] / [800 – (-10000)]

= 54000 / 10800

= 5

Then for z:

z = [(AE-DB)(AL-JG)-(AK-DK)(HA-BG)] / [(AE-DB)(IA-GC)-(AF-DC)(HA-BG)]

= [(300-50)(1840-1800)-(3580-2250)(80-40)] / [(300-50)(40-80) – (120-100)(80-40)]

= -43200/10800

= 4

We have now completed the general formula and also managed to prove that it works. Yet, we have not managed to establish a full understanding of this problem until we have managed to solve this problem using another method. Since we have already done this problem using elimination, the only method that is left to attempt with this general formula lies in matrices.

General Boating Problem – Matrices

Firstly, like before, we have to write the same problem in matrix form, except this time – the numbers of the problem have been instead substituted with letters which represent numbers of the problem.

[A B C] [x] = [J]

[D E F] [y] = [K]

[G H I] [z] = [L]

A X = B

A-1X=B

Firstly, we determine the determinant of A-1 by rewriting the Matrix A as:

[A] = A [E F] -B [D F] +C [D E]

[H I ] [G I] [G H]

We then multiply everything out and the determinant is:

A(EI-HF)-B(DI-FG)+C(DH-EG)

We then move onto the next stage and find A transpose by rewriting its rows as columns and then finding the co-factors:

AT = [A D G]

[B E H]

[C F I ]

Ajugate A = [(EI-FH) (CH-BI) (BF-CE)] [J]

[(FG-DI) (AI-CG) (CD-AF)] [K]

[(DH-EG) (BG-AH) (AE-BD)] [L]

Note: negative (BI-CH) has simply been multiplied by -1 and changed to (CH-BI) as have with other signs that need to be inverted. Now we simple multiply J, K and L into the brackets and create a solution of:

x = 1/ A(EI-HF)-B(DI-FG)+C(DH-EG) X [J (EI-FH)+K (CH-BI)+L(BF-CE)]

y = 1/ A(EI-HF)-B(DI-FG)+C(DH-EG) X [J (FG-DI)+K (AI-CG)+L(CD-AF)]

z = 1/ A(EI-HF)-B(DI-FG)+C(DH-EG) X [J (DH-EG)+K (BG-AH)+L(AE-BD)]

Now the general formula has also been solved using the alternative method of Matrices. And hence provided an alternative method to solve a three variable simultaneous set of equations.

Conclusion:

I believe that overall, I have managed to investigate this problem as much as I can at this stage. I have solved the problem using two methods, investigated new methods of mathematics that I had never encountered before and also produced a mathematical formula that will solve almost any simultaneous equation with three pieces of information and three variables. However – I have noticed that there is a flaw to the general formula. Should the denominator to any of the general formulas = 0. And since it is impossible to divide by 0, there would be no solution to the problem using the general formula and the answer would have to be located using an alternative method. From this coursework, I believe I have learned a lot and it feels highly satisfying to finish it in the end.