Business Mathematics - I

Linear functions describe straight line relationships, and much of business mathematics starts here: demand, supply, cost, and revenue are all modeled as lines before harder curves appear. This unit builds the line itself, applies it to markets and budgets, and ends with fitting a line to real data.

Slope and forms of a straight line

The slope m of a line measures how fast y changes when x changes by one unit. Through points $(x_1, y_1)$ and $(x_2, y_2)$:

The same line can be written in several forms. The slope intercept form $y = mx + c$ shows slope m and y intercept c directly. The point slope form $y - y_1 = m(x - x_1)$ builds the line from one known point and the slope. The two point form computes m from two known points with the formula above, then uses the point slope form. The intercept form $\frac{x}{a} + \frac{y}{b} = 1$ shows the x intercept a and y intercept b directly.

Example: the line through $(1, 5)$ and $(3, 11)$ has slope $m = (11 - 5)/(3 - 1) = 3$, so $y - 5 = 3(x - 1)$, or $y = 3x + 2$. Check: at $x = 3$, $y = 3(3) + 2 = 11$, which matches.

Linear demand, supply, cost, and revenue

A linear demand function links price P to quantity demanded $Q_d$ with a negative slope. That negative slope is the Law of Demand: quantity falls as price rises.

A linear supply function has a positive slope: higher prices call out more quantity.

FIGURE NEEDED: linear demand and supply curves on price quantity axes, showing the downward and upward slopes.

On the firm side, total cost adds fixed cost FC to variable cost v per unit for output Q:

Total revenue is price times quantity, $TR = P \times Q$. When the selling price is fixed, revenue is also a straight line through the origin with slope P.

Elasticity on linear functions

Point price elasticity of demand measures the percentage change in quantity per one percent change in price:

For the linear demand curve $Q_d = a - bP$, the derivative $dQ/dP$ equals $-b$, so elasticity at any point is $E_p = b(P/Q)$. Note that elasticity changes along the line even though the slope is constant. Example: for $Q_d = 100 - 2P$ at $P = 20$, quantity is $Q = 60$, so $E_p = 2(20/60) \approx 0.67$, which is inelastic since it is below 1. Supply elasticity uses the same formula without the minus sign and is positive. Income elasticity replaces price with income Y, $E_Y = (dQ/dY)(Y/Q)$: a positive value marks a normal good and a negative value marks an inferior good.

Budget and cost constraints

A budget line shows all combinations of two goods that exactly spend income M at prices $P_1$ and $P_2$:

Its intercepts are $M/P_1$ and $M/P_2$, and its slope is $-P_1/P_2$. Example: with $M = 1000$, $P_1 = 50$, and $P_2 = 100$, the budget line $50X + 100Y = 1000$ meets the axes at $X = 20$ and $Y = 10$. An isocost (cost constraint) line $C = wL + rK$ works the same way for labour L and capital K at prices w and r.

Least squares line of best fit

Given scattered data on two variables, the least squares line $\hat{y} = a + bx$ is the line that minimizes the sum of squared vertical deviations $\sum (y - \hat{y})^2$. Its coefficients solve the normal equations:

Trend fitting for time series data

For yearly data, the same method fits a trend line by coding time so that $\sum x = 0$, which reduces the normal equations to $a = \sum y / n$ and $b = \sum xy / \sum x^2$. For an odd number of years, set the middle year to $x = 0$. For an even number of years, center the origin between the two middle years with $x = \dots, -3, -1, 1, 3, \dots$. Always shift the origin this way: using raw calendar years as x produces huge numbers and arithmetic errors. Forecasts come from substituting the future coded time value into $\hat{y} = a + bx$. The past question bank asks 10 mark trend fitting questions on yearly income and profit tables, so this procedure is worth practicing with full tables.

Worked problem: trend line and forecast

Data: yearly profits of a trading firm in Rs. millions.

Fit a straight line trend by least squares and estimate profit for 2026.

Solution: with $n = 7$ years, set the origin at the middle year 2020 so $x = t - 2020$.

Since $\sum x = 0$:

For 2026, $x = 6$: