The problem consists of three questions. Q-1 (a): A person receives a pension of Tk. 1400 per year, paid in half-yearly installments. The expectation of life is 15 years, and the interest rate is 4% per annum, payable half-yearly. We need to find the single sum equivalent to this pension. Q-1 (b): A machine costs Tk. 98,000 and has an effective life of 12 years. The scrap value is Tk. 3,000. We need to calculate the amount to be retained out of profits at the end of each year to accumulate at compound interest at 5% per annum. Q-2: A trucking company has three types of trucks (X, Y, Z) that carry three types of machines (I, II, III). The number of each type of machine per truck type is given in a table. We need to estimate how many trucks of each type are needed to carry exactly 29 of machine type I, 13 of machine type II, and 16 of machine type III, using the matrix inverse method. Q-3 (a): The demand function is $P = 500 - 0.2x$, and the cost function is $C = 25x + 10000$. We need to determine the output at which profits are maximum and the price the firm will charge. Q-3 (b): Show that the function $y = x^3 - 3x^2$ has a point of inflexion at the point (1, -2).

1. Problem Description

The problem consists of three questions.

Q-1 (a): A person receives a pension of Tk. 1400 per year, paid in half-yearly installments. The expectation of life is 15 years, and the interest rate is 4% per annum, payable half-yearly. We need to find the single sum equivalent to this pension.

Q-1 (b): A machine costs Tk. 98,000 and has an effective life of 12 years. The scrap value is Tk. 3,

0

0

0. We need to calculate the amount to be retained out of profits at the end of each year to accumulate at compound interest at 5% per annum.

Q-2: A trucking company has three types of trucks (X, Y, Z) that carry three types of machines (I, II, III). The number of each type of machine per truck type is given in a table. We need to estimate how many trucks of each type are needed to carry exactly 29 of machine type I, 13 of machine type II, and 16 of machine type III, using the matrix inverse method.

Q-3 (a): The demand function is $P = 500 - 0.2x$, and the cost function is $C = 25x + 10000$. We need to determine the output at which profits are maximum and the price the firm will charge.

Q-3 (b): Show that the function $y = x^3 - 3x^2$ has a point of inflexion at the point (1, -2).

2. Solution Steps

Q-1 (a):

The pension is Tk. 1400 per year, paid in half-yearly installments, so the installment is Tk.

7

0

0. The interest rate is 4% per annum, payable half-yearly, so the interest rate per half-year is 2% or 0.

0

2. The number of periods is 15 years * 2 =

3

0. We need to find the present value of an annuity of Tk. 700 for 30 periods at an interest rate of 2%.

The formula for the present value of an annuity is:

$PV = PMT * \frac{1 - (1 + r)^{-n}}{r}$

Where:

$PV$ is the present value

$PMT$ is the periodic payment (700)

$r$ is the interest rate per period (0.02)

$n$ is the number of periods (30)

$PV = 700 * \frac{1 - (1 + 0.02)^{-30}}{0.02}$

$PV = 700 * \frac{1 - (1.02)^{-30}}{0.02}$

$PV = 700 * \frac{1 - 0.55207}{0.02}$

$PV = 700 * \frac{0.44793}{0.02}$

$PV = 700 * 22.3965$

$PV = 15677.55$

Q-1 (b):

The cost of the machine is Tk. 98,000, and the scrap value is Tk. 3,

0

0

0. The depreciable amount is Tk. 98,000 - Tk. 3,000 = Tk. 95,

0

0

0. We need to find the annual amount (A) that should be retained to accumulate Tk. 95,000 in 12 years at 5% interest per annum.

The formula for the future value of an ordinary annuity is:

$FV = A * \frac{(1 + r)^n - 1}{r}$

Where:

$FV$ is the future value (95000)

$A$ is the annual amount

$r$ is the interest rate (0.05)

$n$ is the number of years (12)

$95000 = A * \frac{(1 + 0.05)^{12} - 1}{0.05}$

$95000 = A * \frac{(1.05)^{12} - 1}{0.05}$

$95000 = A * \frac{1.79586 - 1}{0.05}$

$95000 = A * \frac{0.79586}{0.05}$

$95000 = A * 15.9171$

$A = \frac{95000}{15.9171}$

$A = 5968.52$

Q-2:

Let $x$ be the number of trucks of type X, $y$ be the number of trucks of type Y, and $z$ be the number of trucks of type Z.

We can set up the following system of equations:

$2x + 3y + 4z = 29$ (Machine I)

$1x + 1y + 2z = 13$ (Machine II)

$3x + 2y + 1z = 16$ (Machine III)

In matrix form, $AX = B$, where:

$A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & 1 & 2 \\ 3 & 2 & 1 \end{bmatrix}$

$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$B = \begin{bmatrix} 29 \\ 13 \\ 16 \end{bmatrix}$

To find $X$, we need to calculate $A^{-1}$ and then $X = A^{-1}B$.

First, we find the determinant of A:

$det(A) = 2(1*1 - 2*2) - 3(1*1 - 2*3) + 4(1*2 - 1*3) = 2(-3) - 3(-5) + 4(-1) = -6 + 15 - 4 = 5$

Now, we find the adjugate (transpose of the cofactor matrix) of A:

Cofactors:

$C_{11} = (1*1 - 2*2) = -3$

$C_{12} = -(1*1 - 2*3) = 5$

$C_{13} = (1*2 - 1*3) = -1$

$C_{21} = -(3*1 - 4*2) = 5$

$C_{22} = (2*1 - 4*3) = -10$

$C_{23} = -(2*2 - 3*3) = 5$

$C_{31} = (3*2 - 4*1) = 2$

$C_{32} = -(2*2 - 4*1) = 0$

$C_{33} = (2*1 - 3*1) = -1$

$adj(A) = \begin{bmatrix} -3 & 5 & 2 \\ 5 & -10 & 0 \\ -1 & 5 & -1 \end{bmatrix}$

$A^{-1} = \frac{1}{det(A)} * adj(A) = \frac{1}{5} \begin{bmatrix} -3 & 5 & 2 \\ 5 & -10 & 0 \\ -1 & 5 & -1 \end{bmatrix} = \begin{bmatrix} -0.6 & 1 & 0.4 \\ 1 & -2 & 0 \\ -0.2 & 1 & -0.2 \end{bmatrix}$

Now, we calculate $X = A^{-1}B$:

$X = \begin{bmatrix} -0.6 & 1 & 0.4 \\ 1 & -2 & 0 \\ -0.2 & 1 & -0.2 \end{bmatrix} \begin{bmatrix} 29 \\ 13 \\ 16 \end{bmatrix} = \begin{bmatrix} -0.6*29 + 1*13 + 0.4*16 \\ 1*29 - 2*13 + 0*16 \\ -0.2*29 + 1*13 - 0.2*16 \end{bmatrix} = \begin{bmatrix} -17.4 + 13 + 6.4 \\ 29 - 26 + 0 \\ -5.8 + 13 - 3.2 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \\ 4 \end{bmatrix}$

So, $x = 2$, $y = 3$, $z = 4$.

Q-3 (a):

The demand function is $P = 500 - 0.2x$, and the cost function is $C = 25x + 10000$.

Revenue, $R = P*x = (500 - 0.2x)x = 500x - 0.2x^2$

Profit, $\pi = R - C = (500x - 0.2x^2) - (25x + 10000) = 475x - 0.2x^2 - 10000$

To maximize profit, we take the derivative with respect to x and set it to 0:

$\frac{d\pi}{dx} = 475 - 0.4x = 0$

$0.4x = 475$

$x = \frac{475}{0.4} = 1187.5$

Since the output must be an integer, we check $x = 1187$ and $x = 1188$.

However, we can accept 1187.5 as the answer for this exam.

To confirm that this is a maximum, we take the second derivative:

$\frac{d^2\pi}{dx^2} = -0.4$

Since the second derivative is negative, this is a maximum.

The output at which profits are maximum is $x = 1187.5$.

The price it will charge is $P = 500 - 0.2(1187.5) = 500 - 237.5 = 262.5$

Q-3 (b):

$y = x^3 - 3x^2$

To find the point of inflexion, we need to find the second derivative and set it to

0. First derivative:

$y' = 3x^2 - 6x$

Second derivative:

$y'' = 6x - 6$

Set $y'' = 0$:

$6x - 6 = 0$

$6x = 6$

$x = 1$

When $x = 1$, $y = (1)^3 - 3(1)^2 = 1 - 3 = -2$

So the point is (1, -2).

To confirm that this is a point of inflexion, we need to check the sign of $y''$ around $x=1$.

When $x = 0$, $y'' = -6 < 0$

When $x = 2$, $y'' = 6 > 0$

Since the sign of $y''$ changes around $x = 1$, (1, -2) is a point of inflexion.

3. Final Answer

Q-1 (a): The single sum equivalent to the pension is Tk. 15677.

5

5. Q-1 (b): The amount that should be retained out of profits at the end of each year is Tk. 5968.

5

2. Q-2: 2 trucks of type X, 3 trucks of type Y, and 4 trucks of type Z are needed.

Q-3 (a): The output at which profits are maximum is 1187.5, and the price it will charge is 262.

5. Q-3 (b): The function $y = x^3 - 3x^2$ has a point of inflexion at (1, -2).