Answer to Question #225797 in Linear Algebra for Himanshu

Question

3(a) An amount of Rs. 8,000 is distributed into three investments at the rate of 7%, 8% and

9% per annum respectively. The total annual income is Rs. 317.50 and the annual income from

the first investment is Rs. 5 more than the income from the second. Find the amount of each

investment using matrix algebra.

3(b) A wholesaler of pencils charges Rs. 24 per dozen on orders of 50 dozens or less. For

orders in excess of 50 dozens, the price is reduced by 20 paise per dozen in excess of 50 dozens.

Find the size of the order that maximizes his total revenue.

3(c) A computer is purchased for Rs. 50,000. It is estimated to depreciate at 10% per annum

every year. Find its scrap value at the end of 12 years.

Answer 1

3(a) Let $x=$ the amount of the first investment, $y=$ the amount of the second investment, and $z=$ the amount of the third investment.  Then

$$x+y+z=8000$$

$$0.07x+0.08y+0.09z=317.50$$

$$0 .07x=0.08y+5$$

$$\begin{matrix} x+y+z=8000 \\ 7x+8y+9z=31750 \\ 7x-8y=500 \end{matrix}$$

$$\begin{pmatrix} 1 & 1 & 1 & & 8000 \\ 7 & 8 & 9 & & 31750 \\ 7 & -8 & 0 & & 500 \\ \end{pmatrix}$$

$R_2=R_2-7R_1$

$$\begin{pmatrix} 1 & 1 & 1 & & 8000 \\ 0 & 1 & 2 & & -24250 \\ 7 & -8 & 0 & & 500 \\ \end{pmatrix}$$

$R_3=R_3-7R_1$

$$\begin{pmatrix} 1 & 1 & 1 & & 8000 \\ 0 & 1 & 2 & & -24250 \\ 0 & -15 & -7 & & -55500 \\ \end{pmatrix}$$

$R_1=R_1-R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 32250 \\ 0 & 1 & 2 & & -24250 \\ 0 & -15 & -7 & & -55500 \\ \end{pmatrix}$$

$R_3=R_3+15R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 32250 \\ 0 & 1 & 2 & & -24250 \\ 0 & 0 &23 & & -419250 \\ \end{pmatrix}$$

$R_3=R_3/23$

$$\begin{pmatrix} 1 & 0 & -1 & & 32250 \\ 0 & 1 & 2 & & -24250 \\ 0 & 0 &1 & & -\dfrac{419250}{23}\\ \end{pmatrix}$$

$R_1=R_1+R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & \dfrac{322500}{23} \\ 0 & 1 & 2 & & -24250 \\ 0 & 0 &1 & & -\dfrac{419250}{23}\\ \end{pmatrix}$$

$R_2=R_2-2R_3$

$$\begin{pmatrix} 1 & 0 & 0 & & \dfrac{322500}{23} \\ \\ 0 & 1 & 0 & & \dfrac{280750}{23} \\ \\ 0 & 0 &1 & & -\dfrac{419250}{23}\\ \end{pmatrix}$$

Since we obtain $z=-\dfrac{419250}{23}<0,$ then the problem has no solution.

Suppose that an amount of Rs. 4,000 is distributed into three investments

$$\begin{matrix} x+y+z=4000 \\ 7x+8y+9z=31750 \\ 7x-8y=500 \end{matrix}$$

$$\begin{pmatrix} 1 & 1 & 1 & & 4000 \\ 7 & 8 & 9 & & 31750 \\ 7 & -8 & 0 & & 500 \\ \end{pmatrix}$$

$R_2=R_2-7R_1$

$$\begin{pmatrix} 1 & 1 & 1 & & 4000 \\ 0 & 1 & 2 & & 3750 \\ 7 & -8 & 0 & & 500 \\ \end{pmatrix}$$

$R_3=R_3-7R_1$

$$\begin{pmatrix} 1 & 1 & 1 & & 4000 \\ 0 & 1 & 2 & & 3750 \\ 0 & -15 & -7 & & -27500 \\ \end{pmatrix}$$

$R_1=R_1-R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 250 \\ 0 & 1 & 2 & &3750 \\ 0 & -15 & -7 & & -27500 \\ \end{pmatrix}$$

$R_3=R_3+15R_2$

$$\begin{pmatrix} 1 & 0 & -1 & & 250 \\ 0 & 1 & 2 & & 3750 \\ 0 & 0 &23 & & 28750 \\ \end{pmatrix}$$

$R_3=R_3/23$

$$\begin{pmatrix} 1 & 0 & -1 & & 250 \\ 0 & 1 & 2 & & 3750 \\ 0 & 0 &1 & & 1250\\ \end{pmatrix}$$

$R_1=R_1+R_3$

$$\begin{pmatrix} 1 & 0 & 0 & &1500 \\ 0 & 1 & 2 & & 3750 \\ 0 & 0 &1 & & 1250\\ \end{pmatrix}$$

$R_2=R_2-2R_3$

$$\begin{pmatrix} 1 & 0 & 0 & &1500 \\ 0 & 1 & 0 & &1250 \\ 0 & 0 &1 & & 1250\\ \end{pmatrix}$$

$x=1500, y=1250, z=1250.$

We have correct solution.

3(b) Let $x=$the size of the order (the number of dozen).

Total revenue is

$$R=R(x)= \begin{cases} 24x & x\leq50 \\ x(24-0.2(x-50)) &x\geq50 \end{cases}$$

$$R=R(x)= \begin{cases} 24x & x\leq50 \\ 34x-0.2x^2 &x\geq50 \end{cases}$$

For maximum $R$ we have $x=-\dfrac{34}{2(-0.2)}=85>50$

Check $p=24-0.2(85-50)=17>0.$

Order in $85$ dozen maximizes his total revenue.

3(c) Cost price of the computer $A_0=50000.$

Depreciation $r=10\%$

$$A(t)=A_0(1-\dfrac{r}{100})^t$$

Given $t=12\ years$

$$A(12)=50000(1-\dfrac{10}{100})^{12}=14121.48$$

Scrap value at the end of 12 years is Rs. $14121.48.$