Business Mathematics II

$A=\int_0^2 \frac32 x^2\,dx=\left[\frac{x^3}{2}\right]_0^2=\frac{8}{2}=4$

Area = 4 square units.

Write as $4x+0y+6z=100$, $3x+6y+z=100$, $3x+4y+3z=100$.

$D=\begin{vmatrix}4&0&6\\3&6&1\\3&4&3\end{vmatrix}=4(18-4)-0+6(12-18)=56-36=20$

$D_x=\begin{vmatrix}100&0&6\\100&6&1\\100&4&3\end{vmatrix}=100\,[1(18-4)+6(4-6)]=100(2)=200$

$D_y=\begin{vmatrix}4&100&6\\3&100&1\\3&100&3\end{vmatrix}=100\,[4(3-1)-1(9-3)+6(0)]=100(2)=200$

$D_z=\begin{vmatrix}4&0&100\\3&6&100\\3&4&100\end{vmatrix}=100\,[4(6-4)+1(12-18)]=100(2)=200$

$x=\frac{200}{20}=10,\quad y=10,\quad z=10$

Check: $4(10)+6(10)=100$ and $30+60+10=100$.

Under the Bertrand model each firm reacts to the rival's previous price. Substituting Y's reaction into X's:

$P_t^X=45+0.8P_{t-1}^Y=45+0.8\,(45+0.8P_{t-2}^X)$

$P_t^X=81+0.64P_{t-2}^X$

This is the difference equation for $P^X$ (two periods apart). For even $t=2n$, let $p_n=P^X_{2n}$, so $p_n=0.64p_{n-1}+81$.

Equilibrium: $\bar p=\frac{81}{1-0.64}=225$. With $p_0=300$:

$p_n=225+75(0.64)^n$

For $t=10$, $n=5$: $0.64^5=0.10737$, so

$P_{10}^X=225+75(0.10737)\approx 233.05$

$y=(x-1)(x-2)$, so the curve lies below the $x$-axis for $1<x<2$ and above it for $2<x<4$. The area is therefore found in two parts.

$F(x)=\frac{x^3}{3}-\frac{3x^2}{2}+2x,\quad F(1)=\frac56,\ F(2)=\frac23,\ F(4)=\frac{16}{3}$

$\int_1^2 y\,dx=\frac23-\frac56=-\frac16\quad(\text{area }\tfrac16)$

$\int_2^4 y\,dx=\frac{16}{3}-\frac23=\frac{14}{3}$

$\text{Area}=\frac16+\frac{14}{3}=\frac{29}{6}\approx4.83\text{ square units}$

$D=\begin{vmatrix}1&1&1\\2&-1&3\\4&5&10\end{vmatrix}=1(-10-15)-1(20-12)+1(10+4)=-25-8+14=-19$

$D_x=\begin{vmatrix}5&1&1\\4&-1&3\\13&5&10\end{vmatrix}=5(-25)-1(40-39)+1(20+13)=-125-1+33=-93$

$D_y=\begin{vmatrix}1&5&1\\2&4&3\\4&13&10\end{vmatrix}=1(40-39)-5(20-12)+1(26-16)=1-40+10=-29$

$D_z=\begin{vmatrix}1&1&5\\2&-1&4\\4&5&13\end{vmatrix}=1(-13-20)-1(26-16)+5(10+4)=-33-10+70=27$

$x=\frac{93}{19},\quad y=\frac{29}{19},\quad z=-\frac{27}{19}$

Check: $x+y+z=\frac{95}{19}=5$ and $2x-y+3z=\frac{186-29-81}{19}=4$.

In the Bertrand model each firm sets its price in response to the rival's price in the previous period. The two reaction functions are identical, so the system is symmetric:

$P_{1,t} = 500 + 0.65P_{2,t-1},\qquad P_{2,t} = 500 + 0.65P_{1,t-1}$

With symmetric firms, $P_2$ follows the same path as $P_1$, and the difference equation is

$P_t = 500 + 0.65P_{t-1}\quad\text{or}\quad P_t - 0.65P_{t-1} = 500$

Solution. The equilibrium (particular solution) is $\bar P = \dfrac{500}{1 - 0.65} = 1428.57$. The general solution is

$P_t = 1428.57 + A(0.65)^t$

With $P_0 = 2500$, $A = 2500 - 1428.57 = 1071.43$, so

$P_t = 1428.57 + 1071.43(0.65)^t$

Period 8: $(0.65)^8 = 0.03186$, so

$P_8 = 1428.57 + 1071.43(0.03186) = 1428.57 + 34.14 \approx 1462.7$

Since $|0.65| < 1$ the path converges to the stable equilibrium price of about 1428.57.

$\frac{x+2}{x-5}=1+\frac{7}{x-5}$

$\int \frac{x+2}{x-5}\,dx = x + 7\ln|x-5| + C$

$f_x = 6xy^2 + 10xy$, so

$f_{xx} = 6y^2 + 10y$

$A^2=\begin{bmatrix}1&2\\3&1\end{bmatrix}\begin{bmatrix}1&2\\3&1\end{bmatrix}=\begin{bmatrix}1+6&2+2\\3+3&6+1\end{bmatrix}=\begin{bmatrix}7&4\\6&7\end{bmatrix}$

Value of the determinant:

$1(15-16)-2(10-12)+3(8-9)=-1+4-3=0$

The matrix is singular (determinant = 0).

For viability, the matrix $I-A$ must satisfy the Hawkins-Simon conditions: all leading principal minors must be positive.

$I-A=\begin{bmatrix}0.2&-0.2\\-0.9&0.3\end{bmatrix}$

• First minor: $0.2>0$ (satisfied).
• Second minor: $|I-A|=0.2(0.3)-(-0.2)(-0.9)=0.06-0.18=-0.12<0$ (not satisfied).

Hence the Hawkins-Simon conditions fail and the system is not viable.

The highest derivative is $\frac{dy}{dx}$, so the order is 1. The highest power of that derivative is 5, so the degree is 5.

Order = 1, Degree = 5.

$y=\int \frac1x\,dx=\ln|x|+C$

The characteristic form is $y_{t+1}=2y_t$, so $y_t=A\,2^t$. With $y_0=3$, $A=3$.

$y_t = 3\cdot 2^t$

$TC=\int(3x^2-10x+3)\,dx=x^3-5x^2+3x+k$

When $x=1$: $1-5+3+k=7$, so $k=8$.

$TC=x^3-5x^2+3x+8$

$AC=\frac{TC}{x}=x^2-5x+3+\frac{8}{x}$

(a) Put $u=\log x$, $du=\frac{dx}{x}$:

$\int\frac{\log x+5}{x}dx=\int(u+5)\,du=\frac{(\log x)^2}{2}+5\log x+C$

(b) Put $u=5x+2$, so $x=\frac{u-2}{5}$, $2x+1=\frac{2u+1}{5}$, $dx=\frac{du}{5}$:

$\int(2x+1)\sqrt{5x+2}\,dx=\frac1{25}\int(2u+1)u^{1/2}du=\frac1{25}\left[\frac45u^{5/2}+\frac23u^{3/2}\right]+C$

$=\frac{4}{125}(5x+2)^{5/2}+\frac{2}{75}(5x+2)^{3/2}+C$

Equilibrium: $8-2x=2x\Rightarrow x_0=2$, $P_0=4$.

Consumer's surplus

$CS=\int_0^2(8-2x)\,dx-x_0P_0=\left[8x-x^2\right]_0^2-8=12-8=4$

Producer's surplus

$PS=x_0P_0-\int_0^2 2x\,dx=8-\left[x^2\right]_0^2=8-4=4$

Consumer's surplus = Rs 4 and producer's surplus = Rs 4 (in money units of price times quantity).

$Q=100L^{0.5}K^{0.5}$, so $Q_L=50\sqrt{K/L}$ and $Q_K=50\sqrt{L/K}$.

$dQ=Q_L\,dL+Q_K\,dK$

With $dL=-0.03L$ and $dK=0.05K$:

$dQ=50\sqrt{K/L}(-0.03L)+50\sqrt{L/K}(0.05K)=\sqrt{LK}\,(-1.5+2.5)=\sqrt{LK}$

Since $Q=100\sqrt{LK}$, $dQ=\frac{Q}{100}$.

Output increases by approximately 1%.

This is linear: $\frac{dy}{dx}+Py=Q$ with $P=\frac1x$, $Q=x^3$.

Integrating factor: $e^{\int dx/x}=x$.

$xy=\int x\cdot x^3\,dx=\frac{x^5}{5}+C$

$y=\frac{x^4}{5}+\frac{C}{x}$

$y_t=C_t+I_t=250+0.9y_{t-1}$

Equilibrium: $\bar y=\frac{250}{1-0.9}=2500$.

With $y_0=3000$:

$y_t=(3000-2500)(0.9)^t+2500=500(0.9)^t+2500$

Stability: the coefficient $0.9$ satisfies $0<0.9<1$, so $(0.9)^t\to0$. The time path decreases smoothly (without oscillation) and converges to the equilibrium income 2500, so it is stable.

Graphical method. Corner points of the feasible region (constraints $2x_1+x_2\le5$, $x_1+x_2\le4$):

• $O(0,0)$: $Z=0$
• $A(2.5,0)$: $Z=12.5$
• $B(0,4)$: $Z=12$
• $C$, intersection of $2x_1+x_2=5$ and $x_1+x_2=4$: subtracting gives $x_1=1$, $x_2=3$; $Z=5(1)+3(3)=14$

The maximum is at $C$.

$x_1=1,\quad x_2=3,\quad Z_{max}=14$

Technical coefficients $a_{ij}=\frac{x_{ij}}{X_j}$ (outputs $X_P=40$, $X_Q=80$):

$A=\begin{bmatrix}16/40&20/80\\8/40&40/80\end{bmatrix}=\begin{bmatrix}0.4&0.25\\0.2&0.5\end{bmatrix}$

$I-A=\begin{bmatrix}0.6&-0.25\\-0.2&0.5\end{bmatrix},\quad |I-A|=0.3-0.05=0.25$

$(I-A)^{-1}=\frac1{0.25}\begin{bmatrix}0.5&0.25\\0.2&0.6\end{bmatrix}=\begin{bmatrix}2&1\\0.8&2.4\end{bmatrix}$

New final demand $F=\begin{bmatrix}18\\44\end{bmatrix}$:

$X=(I-A)^{-1}F=\begin{bmatrix}2(18)+44\\0.8(18)+2.4(44)\end{bmatrix}=\begin{bmatrix}80\\120\end{bmatrix}$

Gross output: P = 80 units, Q = 120 units.

Equilibrium: $\bar y=\frac{21}{1-0.6}=52.5$.

$y_t=(y_0-\bar y)(0.6)^t+\bar y=52.5-2.5(0.6)^t$

$y_1=0.6(50)+21=51$

$y_5=52.5-2.5(0.07776)=52.5-0.1944=52.3056$

So $y_1=51$ and $y_5\approx52.31$.

(a) Excess demand: $Q_d-Q_s=460-25P$.

$\frac{dP}{dt}=0.01(460-25P)=4.6-0.25P\;\Rightarrow\;\frac{dP}{dt}+0.25P=4.6$

Equilibrium $\bar P=\frac{4.6}{0.25}=18.4$. With $P(0)=50$:

$P(t)=18.4+31.6e^{-0.25t}$

For $t=5$: $e^{-1.25}=0.2865$, so $P(5)=18.4+31.6(0.2865)\approx \text{Rs }27.45$.

(b) A drop of Rs 30 means $P=20$:

$18.4+31.6e^{-0.25t}=20\Rightarrow e^{-0.25t}=\frac{1.6}{31.6}=0.05063$

$t=\frac{-\ln(0.05063)}{0.25}=\frac{2.983}{0.25}\approx 11.9$

About 12 time periods.

(a) Revenue and profit.

$TR=xP_x+yP_y=6x-2x^2+25y-0.1y^2$

$\pi=TR-TC=6x-2x^2+25y-0.1y^2-10+6x-5y=12x-2x^2+20y-0.1y^2-10$

(b) Revenue maximization.

$TR_x=6-4x=0\Rightarrow x=1.5$ and $TR_y=25-0.2y=0\Rightarrow y=125$.

Second-order conditions: $TR_{xx}=-4<0$, $TR_{yy}=-0.2<0$, $TR_{xy}=0$, so $TR_{xx}TR_{yy}-TR_{xy}^2=0.8>0$ and it is a maximum.

$TR_{max}=6(1.5)-2(2.25)+25(125)-0.1(15625)=9-4.5+3125-1562.5=1567$

Prices: $P_x=6-2(1.5)=\text{Rs }3$, $P_y=25-0.1(125)=\text{Rs }12.5$.

(c) Profit maximization.

$\pi_x=12-4x=0\Rightarrow x=3$ and $\pi_y=20-0.2y=0\Rightarrow y=100$.

$\pi_{xx}=-4<0$, $\pi_{yy}=-0.2<0$, $\pi_{xx}\pi_{yy}-\pi_{xy}^2=0.8>0$, so this is a maximum. Sell 3 units of X and 100 units of Y.

(d)

$\pi_{max}=12(3)-2(9)+20(100)-0.1(10000)-10=36-18+2000-1000-10=\text{Rs }1008$

Prices: $P_x=6-2(3)=\text{Rs }0$, $P_y=25-0.1(100)=\text{Rs }15$.

$TR=3(0)+100(15)-0=1500$

Check: $TC=10-18+500=492$ and $1500-492=1008$.

$\int\left(3x^4+2x^2+2x+\frac1x\right)dx=\frac{3x^5}{5}+\frac{2x^3}{3}+x^2+\ln|x|+C$

$a_{11}=5,\ a_{12}=8,\ a_{21}=7,\ a_{22}=10$

$A=\begin{bmatrix}5&8\\7&10\end{bmatrix}$

$4(1\cdot7-0\cdot6)-2(3\cdot7-0\cdot9)+5(3\cdot6-1\cdot9)=28-42+45=31$

$f_x=4x^3+8xy^2$, so

$f_{xx}=12x^2+8y^2$

Separate variables: $\frac{dy}{y}=\frac{dx}{2}$. Integrating, $\ln|y|=\frac x2+c$, so

$y=Ce^{x/2}$

$y_t=4y_{t-1}$, so

$y_t=A\cdot4^t$

where $A=y_0$ is an arbitrary constant.

Taking dual variables $u,v$ for the two constraints:

$\text{Maximize } W=6u+2v$

subject to

$2u+v\le5$

$3u+v\le3$

$u,v\ge0$

$I-A=\begin{bmatrix}0.2&-0.7\\-0.6&0.5\end{bmatrix}$

• First minor: $0.2>0$.
• Determinant: $0.2(0.5)-(-0.7)(-0.6)=0.1-0.42=-0.32<0$.

The second leading principal minor is negative, so the Hawkins-Simon condition is not satisfied.

The highest-order derivative is $\frac{d^2y}{dx^2}$, so the order is 2. Its power is 2, so the degree is 2.

Order = 2, Degree = 2.

(a) Put $u=2x+1$, so $2x=u-1$, $dx=\frac{du}{2}$; limits $x=0\to u=1$, $x=2\to u=5$.

$\int_0^2 2x\sqrt{2x+1}\,dx=\frac12\int_1^5(u-1)u^{1/2}du=\frac12\left[\frac25u^{5/2}-\frac23u^{3/2}\right]_1^5$

$=\frac12\left[\left(10\sqrt5-\frac{10\sqrt5}{3}\right)-\left(\frac25-\frac23\right)\right]=\frac{10\sqrt5}{3}+\frac{2}{15}=\frac{50\sqrt5+2}{15}\approx7.587$

(b) Write $x+x\ln x=x(1+\ln x)$ and put $u=1+\ln x$, $du=\frac{dx}{x}$:

$\int\frac{dx}{x(1+\ln x)}=\int\frac{du}{u}=\ln|1+\ln x|+C$

$TR=\int(3x^2+2x+10)\,dx=x^3+x^2+10x$

(the constant is zero because $TR=0$ when $x=0$).

When $x=5$: $TR=125+25+50=\text{Rs }200$.

When $x=10$: $TR=1000+100+100=\text{Rs }1200$.

$TC=\int(12+10x-6x^2)\,dx=12x+5x^2-2x^3+k$

Given $TC(4)=64$: $48+80-128+k=64\Rightarrow k=64$.

$TC=-2x^3+5x^2+12x+64$

(a)

$MPL=\frac{\partial Q}{\partial L}=5L^{-1/2}K^{1/2}=5\sqrt{\frac KL}$

$MPK=\frac{\partial Q}{\partial K}=5L^{1/2}K^{-1/2}=5\sqrt{\frac LK}$

(b) Differentiating $MPL$ with respect to $K$:

$\frac{\partial^2Q}{\partial L\,\partial K}=5\cdot\frac12L^{-1/2}K^{-1/2}=\frac{5}{2\sqrt{LK}}$

This is positive for $L,K>0$, so it is not zero; the stated result $\frac{\partial^2Q}{\partial L\partial K}=0$ does not hold for this production function. (It would hold only if the production function had no interaction between $L$ and $K$, such as $Q=aL+bK$.) The cross partial equals $\frac{\partial^2Q}{\partial K\partial L}$, as Young's theorem requires.

The first equation mixes a derivative with lagged terms, so it is read as the first-order difference equation $y_t-3y_{t-1}=7$.

Equilibrium: $\bar y=\frac{7}{1-3}=-3.5$, so

$y_t=A\cdot3^t-3.5$

For $2y_t-3y_{t-1}=7$ with $y_0=7$, divide by 2:

$y_t-1.5y_{t-1}=3.5\;\Rightarrow\;\bar y=\frac{3.5}{1-1.5}=-7$

$y_t=A(1.5)^t-7,\qquad y_0=7\Rightarrow A=14$

$y_t=14(1.5)^t-7$

Check: $y_1=21-7=14$, and $2(14)-3(7)=7$.

The total cost function is not given in the question, so only the general form can be stated:

$\pi=TR-TC=12x-\frac{x^2}{4}-TC(x)$

For example, with a total cost $TC=c(x)$, profit is maximised where $MR=MC$:

$MR=12-\frac x2$

If costs were, say, $TC=2x+10$, then $\pi=10x-\frac{x^2}{4}-10$, with $x=20$ and $\pi_{max}=90$.

(a) Formulation. Let $x_1$ and $x_2$ be the numbers of watches of models $M_1$ and $M_2$ bought.

$\text{Maximize } Z=70x_1+80x_2$

subject to

$x_1+x_2\le30\quad(\text{showcase})$

$100x_1+200x_2\le4000\;\;(\text{i.e. } x_1+2x_2\le40)\quad(\text{budget})$