Good Mathematics In Management Essay Example

Question #1

Find the straight line that passes through point (2,4) and is perpendicular to line:
2x+12y-6=0

Rewrite the expression of the line in the following form:

12y=-2x+6y=-16x+12

We know that the slope of the line that is perpendicular to the line with slope k is:

k'=-1k

That’s why the slope of this line for our case is:

k'=-1-16=6

And the equation of the perpendicular line is:

y=6x+b

Where b must be determined knowing that the line passes the point (2, 4). Hence:

4=6*2+bb=4-12=-8

The equation of the straight line that passes through point (2,4) and is perpendicular to line given is:

y=6x-8

Given the function

y=3x2-11x+6=0

What is the value of x?

We have to solve the quadratic equation here:
3x2-11x+6=0D=-112-4*3*6=121-72=49D=7x1,2=11±72*3x1=186=3x2=46=23

Answer:

x1=186=3x2=46=23

Question #2

Given the following information:
Total revenue:
TRX=400*50=20000TRY=500*80=40000

Total cost:

TCX=400*10+13+16+2000=17600TCY=500*12+14+18+1000=23000

Total profit:

TPX=TRX-TCX=20000-17600=2400TPY=TRY-TCY=40000-23000=17000

Breakeven point:

For X:
TPX=0=Qx*50-Qx*10+13+16-2000Qx=200011≈181.82

The breakeven point is at 182 units of X.

TPX=182=2

For Y:

TPY=0=QY*80-QY*12+14+18-1000QY=2509≈27.778

The breakeven point is at 28 units of Y.

TPY=28=8
c.

If profit should be 1000, then:

Qx*50-Qx*10+13+16-2000=1000x=300011≈273 units
QY*80-QY*12+14+18-1000=1000QY=5009≈56 units

Question #3

Explain the differences between simple interest and compound interest.
Interest income depends not only on the interest rate, but also on the mechanism of charging interest. For example, if every time accrued interest income, which does not increase the amount of the original contribution, we are talking about simple interest, and when interest capitalization – about compound interest.
If you repeatedly calculating simple interest accrual is made with respect to the initial amount and is every time the same amount. In other words,
where

P is the original amount

S is backfilled amount (initial amount, together with accrued interest)
i is the interest rate expressed as a fraction of the period
n is the number of compounding periods.

In this case we speak of a simple interest rate.

When multiple compound interest accruing each time is relative to the amount already accrued previously percent. In other words,
In this case we speak about compound interest rate.
As it seems from the formulas, the results of S in both cases are different.
If a commodity costs RM1,500 now and inflation is expected to go up at the rate of 9% per annum, how much will the commodity cost in 7 years?

This is compound interest. We have to find future value of commodity cost if it is growing on 9% each year:

FV=1500*1+0.097≈2742.06

The commodity cost in 7 years will be 2,742.06

If you can sell an asset for RM11,000 in 8 years and the cost of money is 5%, how much would you be willing to pay for the asset today?

We have to find the present value of RM11,000:

PV=110001+0.058≈7,445.23
You will be willing to pay RM7,445.23 today for this asset.

References

van Deventer, Donald R. and Kenji Imai (2003). Credit Risk Models and the Basel Accords. John Wiley & Sons. ISBN 978-0-470-82091-9.
Kellison, Stephen G. (1970). The Theory of Interest. Richard D. Irwin, Inc. Library of Congress Catalog Card No. 79-98251.
Sullivan, Arthur; Steven M. Sheffrin (2003). Economics: Principles in action. Upper Saddle River, New Jersey 07458: Pearson Prentice Hall. p. 261. ISBN 0-13-063085-3.