Business Mathematics I

Points $(2,5)$ and $(3,1)$.

Slope:

$m = \frac{1-5}{3-2} = -4$

Using $y - y_1 = m(x - x_1)$:

$y - 5 = -4(x - 2) \Rightarrow y = -4x + 13$

So the equation is $4x + y - 13 = 0$.

Demand: $P = 40 - 0.2Q$. At $P = 20$: $20 = 40 - 0.2Q$, so $Q = 100$.

From the demand function, $Q = 200 - 5P$, so $\dfrac{dQ}{dP} = -5$.

$E_d = \left|\frac{dQ}{dP}\cdot\frac{P}{Q}\right| = 5 \times \frac{20}{100} = 1$

The price elasticity of demand is $1$ (unit elastic).

Sum of roots $= \sqrt5 + (-\sqrt5) = 0$. Product of roots $= \sqrt5 \times (-\sqrt5) = -5$.

The equation is $x^2 - (\text{sum})x + \text{product} = 0$:

$x^2 - 0\cdot x - 5 = 0 \Rightarrow x^2 - 5 = 0$

$2^{x-2}\times 8^x = 16+5 = 21$

Since $8^x = 2^{3x}$:

$2^{x-2+3x} = 2^{4x-2} = 21$

Taking logarithms:

$(4x-2)\log 2 = \log 21 \Rightarrow 4x - 2 = \frac{\log 21}{\log 2} = \frac{1.3222}{0.3010} = 4.3923$

$x = \frac{6.3923}{4} \approx 1.598$

For the A.P. $6, 10, 14, \ldots$: $a = 6$, $d = 4$.

$T_n = a + (n-1)d = 62 \Rightarrow 6 + 4(n-1) = 62 \Rightarrow n - 1 = 14 \Rightarrow n = 15$

So 62 is the 15th term.

Geometric series $1 + 3 + 9 + \ldots$: $a = 1$, $r = 3$, $n = 8$.

$S_n = \frac{a(r^n - 1)}{r - 1} = \frac{3^8 - 1}{3 - 1} = \frac{6561 - 1}{2} = 3280$

$P =$ Rs 12,500, $r = 12\%$, $n = 3$ years.

$A = P(1+r)^n = 12500(1.12)^3 = 12500 \times 1.404928 = \text{Rs } 17{,}561.60$

$CI = A - P = 17561.60 - 12500 = \text{Rs } 5{,}061.60$

$\lim_{x\to2}\frac{x^2-4}{x-2} = \lim_{x\to2}\frac{(x-2)(x+2)}{x-2} = \lim_{x\to2}(x+2) = 4$

$f(x) = x^3 - 12x^2 + 2$

$f'(x) = 3x^2 - 24x, \qquad f''(x) = 6x - 24$

Setting $f''(x) = 0$ gives $x = 4$. Since $f''$ changes sign from negative to positive at $x = 4$ (and $f'''=6\neq0$), it is a point of inflection.

$f(4) = 64 - 192 + 2 = -126$

The point of inflection is $(4, -126)$.

$y = \frac{1}{\sqrt{2x+1}} = (2x+1)^{-1/2}$

$\frac{dy}{dx} = -\frac12(2x+1)^{-3/2}\cdot 2 = -(2x+1)^{-3/2} = -\frac{1}{(2x+1)^{3/2}}$

The equations are:

$x + 2y + 3z = 13 \quad (1)$ $2x + 4y + z = 11 \quad (2)$ $3x + 2y + 2z = 14 \quad (3)$

Using Cramer's rule:

$D = \begin{vmatrix}1&2&3\\2&4&1\\3&2&2\end{vmatrix} = 1(8-2) - 2(4-3) + 3(4-12) = 6 - 2 - 24 = -20$

$D_x = \begin{vmatrix}13&2&3\\11&4&1\\14&2&2\end{vmatrix} = 13(8-2) - 2(22-14) + 3(22-56) = 78 - 16 - 102 = -40$

$D_y = \begin{vmatrix}1&13&3\\2&11&1\\3&14&2\end{vmatrix} = 1(22-14) - 13(4-3) + 3(28-33) = 8 - 13 - 15 = -20$

$D_z = \begin{vmatrix}1&2&13\\2&4&11\\3&2&14\end{vmatrix} = 1(56-22) - 2(28-33) + 13(4-12) = 34 + 10 - 104 = -60$

$x = \frac{D_x}{D} = 2, \qquad y = \frac{D_y}{D} = 1, \qquad z = \frac{D_z}{D} = 3$

Check in (2): $4 + 4 + 3 = 11$. In (3): $6 + 2 + 6 = 14$.

So $x = 2,\ y = 1,\ z = 3$.

Demand: $P = 60 - 0.6Q$. Supply: $P = 20 + 0.2Q$.

(a) Equilibrium

$60 - 0.6Q = 20 + 0.2Q \Rightarrow 40 = 0.8Q \Rightarrow Q_0 = 50$

$P_0 = 20 + 0.2(50) = 30$

Equilibrium quantity is $50$ units and equilibrium price is Rs $30$.

(b) Surpluses

$CS = \int_0^{50}(60 - 0.6Q)\,dQ - P_0Q_0 = \left[60Q - 0.3Q^2\right]_0^{50} - 1500 = (3000 - 750) - 1500 = 750$

$PS = P_0Q_0 - \int_0^{50}(20 + 0.2Q)\,dQ = 1500 - \left[20Q + 0.1Q^2\right]_0^{50} = 1500 - (1000 + 250) = 250$

Consumer surplus $=$ Rs 750 and producer surplus $=$ Rs 250.

(c) Total surplus

$TS = CS + PS = 750 + 250 = \text{Rs } 1{,}000$

$y = x^2 - 5x + 6$

This is an upward-opening parabola ($a = 1 > 0$).

• Roots (x-intercepts): $x^2 - 5x + 6 = (x-2)(x-3) = 0$, so $x = 2$ and $x = 3$.
• y-intercept: at $x = 0$, $y = 6$.
• Vertex: $x = -\dfrac{b}{2a} = \dfrac{5}{2} = 2.5$, and $y = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25$.
• Axis of symmetry: $x = 2.5$.

Plot these points and join them with a smooth U-shaped curve: it starts high at $(0, 6)$, falls through $(1,2)$ and $(2,0)$ to the lowest point $(2.5, -0.25)$, then rises through $(3,0)$, $(4,2)$ and $(5,6)$.

Minimum value: since the parabola opens upward, the minimum occurs at the vertex. Check by calculus: $\dfrac{dy}{dx} = 2x - 5 = 0 \Rightarrow x = 2.5$ and $\dfrac{d^2y}{dx^2} = 2 > 0$.

$y_{\min} = -0.25 = -\frac14$

$V = 250000\,e^{-0.6t}$

After 5 years: $V = 250000\,e^{-3} = 250000 \times 0.049787 \approx \text{Rs } 12{,}446.77$

After 10 years: $V = 250000\,e^{-6} = 250000 \times 0.0024788 \approx \text{Rs } 619.69$

After 15 years: $V = 250000\,e^{-9} = 250000 \times 0.00012341 \approx \text{Rs } 30.85$

After 20 years: $V = 250000\,e^{-12} = 250000 \times 0.0000061442 \approx \text{Rs } 1.54$

The resale value decays rapidly towards zero.

$P = 5000$, $A = 20000$, $r = 5.5\% = 0.055$.

$A = P(1+r)^n \Rightarrow 20000 = 5000(1.055)^n \Rightarrow (1.055)^n = 4$

Taking logarithms:

$n = \frac{\log 4}{\log 1.055} = \frac{0.60206}{0.023252} \approx 25.89$

The sum will grow to Rs 20,000 in about $25.9$ years, i.e. about 26 years.

Price of house $=$ Rs 5,000,000; paid immediately $=$ Rs 2,000,000.

Balance to be paid in instalments: $A = 5{,}000{,}000 - 2{,}000{,}000 =$ Rs 3,000,000, with $i = 0.15$ and $n = 15$.

For equal annual instalments (ordinary annuity) $R$:

$A = R\cdot\frac{1 - (1+i)^{-n}}{i} \Rightarrow R = \frac{A\,i}{1-(1.15)^{-15}}$

$(1.15)^{15} = 8.13706$, so $(1.15)^{-15} = 0.122894$ and $1 - 0.122894 = 0.877106$.

$R = \frac{3{,}000{,}000 \times 0.15}{0.877106} = \frac{450{,}000}{0.877106} \approx \text{Rs } 513{,}051$

Mr. Kumar must pay about Rs 513,051 annually.

(i) $x^2 + y = 9 \Rightarrow y = 9 - x^2$

$\frac{dy}{dx} = -2x$

(ii) $y = t^4 + 2$ and $x = t^3 + 1$ (parametric form)

$\frac{dy}{dt} = 4t^3, \qquad \frac{dx}{dt} = 3t^2$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t^3}{3t^2} = \frac{4t}{3}$

Fit the straight line $Y = a + bX$ by least squares.

There are $n = 7$ (odd) years, so take the middle year 2020 as origin: $X = \text{Year} - 2020$, so $\sum X = 0$.

Normal equations (with $\sum X = 0$):

$\sum Y = na \Rightarrow a = \frac{413}{7} = 59$

$\sum XY = b\sum X^2 \Rightarrow b = \frac{52}{28} = 1.857$

The line of best fit is:

$Y = 59 + 1.857X \quad (X = \text{year} - 2020,\ Y \text{ in Rs millions})$

Estimates

For 2024: $X = 4$, so $Y = 59 + 1.857(4) = 59 + 7.43 = 66.43$.

For 2026: $X = 6$, so $Y = 59 + 1.857(6) = 59 + 11.14 = 70.14$.

Estimated income is about Rs 66.43 million in 2024 and about Rs 70.14 million in 2026.

Expenditure equation

Disposable income: $Y_d = Y - tY = 0.6Y$

Consumption: $C = 90 + 0.8(0.6Y) = 90 + 0.48Y$

Imports: $M = 0.3Y_d = 0.18Y$

Substituting into $Y = C + I + G + X - M$ (all in Rs million):

$Y = 90 + 0.48Y + 500 + 700 + 250 - 0.18Y$

$Y = 1540 + 0.30Y$

This is the expenditure equation.

Equilibrium national income

$0.70Y = 1540 \Rightarrow Y = \frac{1540}{0.70} = \text{Rs } 2200\text{m}$

Equilibrium consumption

$C = 90 + 0.48(2200) = 90 + 1056 = \text{Rs } 1146\text{m}$

Total tax

$T = tY = 0.4 \times 2200 = \text{Rs } 880\text{m}$

Check: $M = 0.18(2200) = 396$, and $1146 + 500 + 700 + 250 - 396 = 2200$, so the equilibrium holds.

Equilibrium with a tax of Rs $t$ per unit

The tax raises the price at which suppliers are willing to sell, so the supply price paid by consumers is $P_s + t$. Setting this equal to the demand price:

$3Q + 40 + t = -2Q + 80 \Rightarrow 5Q = 40 - t \Rightarrow Q = \frac{40 - t}{5}$

Tax revenue

$T = tQ = \frac{t(40 - t)}{5} = \frac{40t - t^2}{5}$

For a maximum, $\dfrac{dT}{dt} = \dfrac{40 - 2t}{5} = 0 \Rightarrow t = 20$.

Also $\dfrac{d^2T}{dt^2} = -\dfrac{2}{5} < 0$, so revenue is maximised at $t = \text{Rs } 20$.

(a) Equilibrium price and quantity

$Q = \frac{40 - 20}{5} = 4 \text{ units}$

Price paid by consumers: $P = -2(4) + 80 = \text{Rs } 72$

(Sellers receive $3(4) + 40 = \text{Rs } 52$, and $72 - 52 = 20 = t$.)

(b) Total tax raised

$T = tQ = 20 \times 4 = \text{Rs } 80$

For a one-year project, $\text{NPV} = \dfrac{R}{1.09} - \text{outlay}$ and the IRR satisfies $\dfrac{R}{1+r} = \text{outlay}$, i.e. $r = \dfrac{R}{\text{outlay}} - 1$.

Project A: $r = \dfrac{11000}{10000} - 1 = 10\%$. $\text{NPV} = \dfrac{11000}{1.09} - 10000 = 10091.74 - 10000 = \text{Rs } 91.74$

Project B: $r = \dfrac{6520}{6000} - 1 = 8.67\%$. $\text{NPV} = \dfrac{6520}{1.09} - 6000 = 5981.65 - 6000 = -\text{Rs } 18.35$

Project C: $r = \dfrac{9800}{9000} - 1 = 8.89\%$. $\text{NPV} = \dfrac{9800}{1.09} - 9000 = 8990.83 - 9000 = -\text{Rs } 9.17$

Project D: $r = \dfrac{8700}{8000} - 1 = 8.75\%$. $\text{NPV} = \dfrac{8700}{1.09} - 8000 = 7981.65 - 8000 = -\text{Rs } 18.35$

Decision: A project is viable if its IRR exceeds the market rate of 9% (equivalently, NPV is positive). Only Project A is viable.

Ranking by profitability (IRR): A (10.00%) > C (8.89%) > D (8.75%) > B (8.67%). Only A should be accepted; the others would earn less than the market rate of interest.

(a) Revenue, cost and profit functions

Revenue: $R = Px = (500 - 2x)x = 500x - 2x^2$

Total variable cost: $VC = x(100 + 2x) = 100x + 2x^2$

Cost: $C = 200 + 100x + 2x^2$

Profit: $\pi = R - C = 500x - 2x^2 - 200 - 100x - 2x^2$

$\pi = 400x - 4x^2 - 200$

(b) Maximum revenue

$R' = 500 - 4x = 0 \Rightarrow x = 125$. Since $R'' = -4 < 0$, this is a maximum.

$R_{max} = 500(125) - 2(125)^2 = 62500 - 31250 = 31250$

Maximum revenue is Rs 31,250 (at price $P = 500 - 250 = 250$).

(c) Maximum profit

$\pi' = 400 - 8x = 0 \Rightarrow x = 50$. Since $\pi'' = -8 < 0$, this is a maximum.

$\pi_{max} = 400(50) - 4(50)^2 - 200 = 20000 - 10000 - 200 = 9800$

Maximum profit is Rs 9,800.

(d) Breakeven points

Set $\pi = 0$: $4x^2 - 400x + 200 = 0 \Rightarrow x^2 - 100x + 50 = 0$

$x = \frac{100 \pm \sqrt{10000 - 200}}{2} = 50 \pm \sqrt{2450} = 50 \pm 49.50$

So $x \approx 0.50$ and $x \approx 99.50$ units. The firm makes a profit between about 0.5 and 99.5 units and a loss outside this range.