Mathematics for Economists: free practice, theory and problems

1. Functions, marginal analysis and elasticity

What is it about?

Economists use functions for cost, revenue and demand. The derivative is called marginal (what happens with one more unit), and elasticity measures how sensitive demand is to price.

Concepts and formulas

• Marginal cost $K'(x)$, marginal revenue $I'(x)$. Maximum profit where $I'(x) = K'(x)$.
• Linear demand: $x = a - bp$. Revenue: $I = p \cdot x$.
• Price elasticity: $E_p = \dfrac{dx}{dp}\cdot\dfrac{p}{x}$. Interpretation: the percentage change in quantity demanded when the price rises by 1 %.
• $|E_p| > 1$: elastic (a price rise gives lower revenue). $|E_p| < 1$: inelastic (a price rise gives higher revenue).
• Revenue is largest where $|E_p| = 1$.

Example

$x = 1000 - 5p$ and $p = 80$: $x = 600$, $\dfrac{dx}{dp} = -5$, $E_p = -5 \cdot \dfrac{80}{600} = -0.67$. Inelastic: a price rise gives more revenue.

Practise functions, marginal analysis and elasticity in the app →

2. Cost, revenue and profit

🔗 Shared curriculum with Mathematics S1. The questions are the same, but your progress counts separately for each subject.

What is it about?

Businesses use functions to describe what it costs to produce $x$ units, what they earn, and when the profit is largest. The derivative is here called marginal cost and marginal revenue.

Concepts and formulas

• Cost function $K(x)$: fixed costs (the constant term) plus variable costs. Example: $K(x) = 0.02x^2 + 20x + 5000$.
• Revenue function at a fixed price $p$: $I(x) = p \cdot x$.
• Profit: $O(x) = I(x) - K(x)$.
• Marginal cost $K'(x)$: roughly what it costs to make one more unit. Marginal revenue $I'(x)$ likewise.
• Largest profit when $O'(x) = 0$, i.e. when marginal revenue = marginal cost: $I'(x) = K'(x)$.
• Unit cost $E(x) = \dfrac{K(x)}{x}$. It is smallest where $E(x) = K'(x)$.

Example

$K(x) = 0.02x^2 + 20x + 5000$ and the price is 100 NOK. $O'(x) = 100 - (0.04x + 20) = 0$ gives $x = 2000$ units. The profit is then $O(2000) = 200\,000 - (80\,000 + 40\,000 + 5000) = 75\,000$ NOK.

Practise cost, revenue and profit in the app →

3. Differentiation rules in practice

🔗 Shared curriculum with Mathematics R1. The questions are the same, but your progress counts separately for each subject.

What is it about?

Here you practise the rules you need to differentiate everything in R1: powers, $e^x$, $\ln x$, and combinations with the product, quotient and chain rules.

Concepts and formulas

• $(x^n)' = nx^{n-1}$, $(e^x)' = e^x$, $(\ln x)' = \dfrac{1}{x}$.
• The chain rule: $\big(e^{kx}\big)' = ke^{kx}$, $\big(\ln u\big)' = \dfrac{u'}{u}$, $\big((ax + b)^n\big)' = na(ax + b)^{n-1}$.
• The product rule: $(uv)' = u'v + uv'$. Example: $(x e^x)' = e^x + xe^x = (1 + x)e^x$.
• The quotient rule: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$. Example: $\left(\dfrac{x}{x + 1}\right)' = \dfrac{1}{(x + 1)^2}$.

Method

1. Look at the structure: is it a product, a quotient or a function of a function?
2. Choose the rule and write down $u$, $v$ (or the inner function $u$) and their derivatives.
3. Put it together and simplify. Insert numbers at the very end.

Practise differentiation rules in practice in the app →

4. Savings, loans and present value

🔗 Shared curriculum with Mathematics S2. The questions are the same, but your progress counts separately for each subject.

What is it about?

Savings, loans and investments are geometric series in disguise. Each payment grows (or is discounted) by the same factor, so the sum becomes a geometric series.

Concepts and formulas

• Growth factor per period: $k = 1 + r$, where $r$ is the interest rate per period as a decimal.
• Savings (future value): $a$ NOK is deposited at the end of each period for $n$ periods. Right after the last deposit the balance is $S = a \cdot \dfrac{k^n - 1}{k - 1}$.
• Present value: an amount $B$ in $n$ periods is worth $\dfrac{B}{k^n}$ today. The present value of $n$ equal amounts $a$ (the first in one period) is $N = a \cdot \dfrac{1 - k^{-n}}{r}$.
• Annuity loan: equal instalments. The loan $L$ is the present value of all instalments, so $a = \dfrac{L \cdot r}{1 - (1 + r)^{-n}}$.
• Serial loan: equal repayments $L/n$. Interest is charged on the remaining loan, so the instalment decreases over time.

Example

A loan of 1,000,000 NOK, 5 % interest, 20 annual instalments (annuity): $a = \dfrac{1\,000\,000 \cdot 0.05}{1 - 1.05^{-20}} \approx 80\,243$ NOK per year. As a serial loan, the first instalment is $50\,000 + 50\,000 = 100\,000$ NOK, and it decreases by 2500 NOK each year.

Practise savings, loans and present value in the app →