• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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

Extracts from this document...

Introduction

Maths Coursework

By Jeffrey Li

The teacher has presented me with a mathematical problem in which I must solve and investigate. It is as follows:

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

...read more.

Middle

[ +  -  +]        

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

...read more.

Conclusion

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.

...read more.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related AS and A Level Core & Pure Mathematics essays

  1. Marked by a teacher

    C3 Coursework - different methods of solving equations.

    5 star(s)

    The chance of this happening is quite high. As for the speed of convergence, it would be slower that Newton Raphson because of the extra time needed to re-arrange the equation but it is faster than Decimal search. With the use of Excel, it is very easy to enter the x = g(x)

  2. The open box problem

    X 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 V 121.5 123.904 125.732 127.008 127.756 128 127.764 127.072 125.948 124.416 122.5 From this table we can see that the maximum volume is 128 and the x is 2.

  1. Best shape for gutter and further alegbra - using Excel to solve some mathematical ...

    best cross section to use for a gutter to carry the maximum water capacity is a semi circle. 2) Using Excel; it will take 21 months to clear the debt. Of which that last month is a small payment under �70.

  2. Numerical Method (Maths Investigation)

    I only got its range, -0.66514 < root of the equation used < -0.66513. So, I decide to get their error bound by: Average of the upper and lower error bound = = Upper Error Bound = Lower Error Bound = The Error Bound of the root of this equation

  1. Numerical solutions of equations

    By looking at the diagram (see Figure 3), I can see that my starting value (x1) is -1.5. x1 = -1.5 x2 = -1.295 (I substituted x1 = -1.5 into the iterative formula to get the value for x2) x3 = -1.227690 (I substituted x2 = -1.295 into the iterative formula to get x3)

  2. Math Portfolio Type II - Applications of Sinusoidal Functions

    = 1.627 sin[0.016(n -111.744)] + 6.248. This is found through using the TI-84 Plus graphing calculator. All the coordinates are listed in a table and the equation is found using sinusoidal regression, which is done with the graphing calculator. 3.

  1. Numerical solution of equations

    Here are three steps of this method: - Take increment in x of size 0.1 within the interval of the root. Stop when there is a sign change. - The interval has been narrowed down and now keeps on taking increment in x of size 0.01 within the narrowed interval.

  2. Functions. Mappings transform one set of numbers into another set of numbers. We could ...

    can order that is equal to or larger than the denominator ? The remainder is written as a fraction ==> There are two methods for doing this ? Polynomial long division ? Remainder theorem ==> Remainder has to have a lower power than the divisor Let F(x)

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work