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ASSIGNMENT

MATHEMATICS

JULY 2014 SEMESTER

KOD KURSUS

NAMA KURSUS

PROGRAM

NAMA PELAJAR

NO. MATRIK

NAMA FASILITATOR

AKADEMIK

PUSAT PEMBELAJARAN

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Question 1

a) Sketch the line with the given equation:

F = 9/5 C + 32

b) What is the slope of the line? What does it represent?

The slope of the line is 9/5 or 1.8.

The slope of a graph illustrates the rate of change from one point on the graph to another

point. It may be either positively sloped, sloping upward from left to right, or negatively

sloped, sloping downward from left to right.

In this question, the slope has shown a positive slope as it move upward from left to right.

Steeper graphs have a greater rate of change than do shallower graphs. A perfectly vertical

graph has an undefined slope, while a perfectly horizontal graph has a slope of zero.

c) What is the F-intercept of the line? What does it represent?

The F-intercept is 9/5C + 32.

The F-intercept is the coefficient value in a linear equation. The F-intercept is important

because it is the point in which you start plotting the line. In this case, the amount C will

affect the determination of F-intercept.

[1]

F

C

F = 9/5 C + 32

Question 2

a) What is the cost of function?

Cost function formula would be, c(x) = ax + b

This is to be said as, Total Cost (TC) = Variable Cost (VC) + Fixed Cost (FC)

x = Units of CD’s

b = FC = RM 30,000.00

a = VC = RM 6.00 per CD

Therefore, the cost function formula would be:

c(x) = 6x + 30,000

b) What is the revenue function?

To find revenue, we need to multiply the price of CD with the Units of CD sold.

R = revenue

X = Units of CD sold

Therefore, the revenue function would be:

R = 10x

c) What is the profit function?

To find profit, we need to minus the Total Cost from The Revenue.

Profit, P = Revenue (R) – Total Cost [c(x)]

Therefore, the profit function would be:

= R – c(x)

= 10x – (6x – 30,000)

= 4x-30,000

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d) Compute the profit (loss) corresponding to production levels of:

(i) 6000 units of CD

P = 4x-30,000

P = 4 (6,000) – 30,000

P = 24,000 – 30,000

P = -6,000.00

Therefore, if the company sells the CDs of 6000 units, it will facing a loss of RM6,000.00.

(ii) 8000 units of CD

P = 4x-30,000

P = 4 (8,000) – 30,000

P = 32,000 – 30,000

P = 2,000.00

Therefore, if the company sells the CDs of 8000 units, it will be having a profit of

RM2,000.00.

(iii) 12000 units of CD

P = 4x-30,000

P = 4 (12,000) – 30,000

P = 48,000 – 30,000

P = 18,000.00

Therefore, if the company sells the CDs of 8000 units, it will be having a profit of

RM18,000.00.

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Question 3

New machine at cost RM 250,000.00

Depreciated for 10 years

Scrap value RM 10,000.00

(a) Find an expression of the machine’s book value in the T year of use.

Y2 = 0

Y1 = 250,000

X2 = 10

X1 = 0

M = (Y2 – Y1) / (X2 – X1)

M = (0 – 250,000) / (10 – 0)

M = -250,000 / 10

M = -25,000

b = 250,000

The expression of the machine book value in the T years is:

v(T) = m(T) + b

v(T) = -25,000T + 250,000

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(b) Sketch the graph of the function of part (a).

(c) Find the machine’s book value in 2011.

The machine was bought on 2007. Therefore, the machine’s book value in 2011 will occur 4 years of period.

v(T) = -25,000(T) + 250,000

v(T) = -25,000 (4) + 250,000

v(T) = -100,000 + 250,000

v(T) = 150,000

Therefore, the machine’s book value in 2011 would be RM 150,000.00.

(d) Find the rate at which the machine is being depreciated.

Depreciation = (RM 250,000 – RM 10,000) / 10 years

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Time (Years)

Value (RM)

RM 250,000.00

RM 100,000.00

10 Years

= RM 240,000 / 10 years

= RM 24,000.00 per year

Therefore, the rate of machine being depreciated would be:

= (RM 24,000.00 / RM 250,000.00) x 100

= 0.096 x 100

= 9.6 % a year

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