Curve Analysis in Economics and Management

Abstract

Analyse curves in economics and management using graphical reading, the equation of a line, and the solving of first-degree systems of equations and inequalities, in order to visualise and interpret situations of decision, constraint, and equilibrium.

Learning objectives

Read information directly from a graph in the Cartesian plane.
Find and interpret the equation of a line.
Solve first-degree systems of equations (graphically and algebraically).
Model and solve first-degree inequalities and systems of inequalities.

1 Diagnostic Activity

Answer these three questions before reading the module. The goal is to surface what you already know and highlight where to pay extra attention.

1. Read the \(y\)-intercept of the line \(y = 2x + 3\).

2. Compute the slope of the line passing through \((0, 1)\) and \((2, 5)\).

3. Is the line \(y = -x + 4\) increasing or decreasing?

Answers (click to reveal)

The \(y\)-intercept is \(\mathbf{3}\) — the constant term in \(y = 2x + 3\).
\(m = \dfrac{5 - 1}{2 - 0} = 2\).
Decreasing, because the slope \(m = -1 < 0\).

Use your results to identify topics needing attention.

2 Opening Problem

Two mobile phone plans are available: Plan A costs €10/month plus €0.05 per minute of calls. Plan B costs €25/month with unlimited calls (€0.00 per minute).

At what number of minutes per month do the two plans cost the same? Below this threshold, which plan is cheaper?

Write down your estimate before continuing.

3 Core Concepts

The Cartesian Plane

A plane is a flat, two-dimensional, unlimited surface. In a plane we can represent, among other things, points and lines. The plane is equipped with an orthonormal coordinate system made of two axes: the horizontal axis of abscissas (\(x\)-axis) and the vertical axis of ordinates (\(y\)-axis). The two axes are perpendicular and the graduations on each axis are equally spaced.

A point is located by a pair of coordinates \((x, y)\): its abscissa \(x\) read on the horizontal axis and its ordinate \(y\) read on the vertical axis.

Pedagogical uses:

reading information directly from a graph;
checking an equation by graphical superposition;
illustrating an economic equilibrium or a break-even point.

The Equation of a Line

A (non-vertical) line has an equation of the form

\[ y = m x + p, \]

where \(m\) is the slope (rate of change of \(y\) with respect to \(x\)) and \(p\) is the \(y\)-intercept (the ordinate where the line crosses the \(y\)-axis, i.e. the value of \(y\) when \(x = 0\)).

Reading the slope from two points

Given two points \(A(x_A, y_A)\) and \(B(x_B, y_B)\) on the line, \[ m = \frac{y_B - y_A}{x_B - x_A}, \qquad p = y_A - m\,x_A. \] A positive slope means the line rises; a negative slope means it falls; a zero slope is a horizontal line.

Worked example — equation of a line

A line passes through \(A(0, 3)\) and \(B(2, 7)\). Then \(m = \dfrac{7 - 3}{2 - 0} = 2\) and, since \(A\) has abscissa \(0\), \(p = 3\). The equation is \(y = 2x + 3\).

Systems of First-Degree Equations

An equation can precisely model economic, financial, or organisational relationships; it lets us compute, explain, forecast, or find equilibrium points. A system of two equations models choices, allocations, or equilibria between two variables — for example:

equilibrium between supply and demand on a market;
allocation of costs between two departments;
pricing of two services;
planning of two activities.

A system of two first-degree equations can be solved in two equivalent ways: graphically or algebraically.

Graphical method

The graphical solution of a system of two equations is the intersection point of the two associated lines. A system has:

one solution if the lines are secant (cross once);
no solution if the lines are parallel and distinct;
infinitely many solutions if the lines coincide.

Derivation — intersection of two lines by substitution (click to expand)

Consider two secant lines \(y = m_1 x + p_1\) and \(y = m_2 x + p_2\) with \(m_1 \neq m_2\). At the intersection the two ordinates are equal, so

\[ m_1 x + p_1 = m_2 x + p_2 \;\Longrightarrow\; (m_1 - m_2)\,x = p_2 - p_1 \;\Longrightarrow\; x = \frac{p_2 - p_1}{m_1 - m_2}. \]

Because \(m_1 \neq m_2\) the denominator is non-zero, confirming a unique solution. The ordinate is then obtained by substituting this \(x\) back into either line: \(y = m_1 x + p_1\). If \(m_1 = m_2\) the lines are parallel: no solution (if \(p_1 \neq p_2\)) or infinitely many (if \(p_1 = p_2\)).

Worked example — algebraic method

Solve \(\begin{cases} 3x + 2y = 12 \\ x - y = 1 \end{cases}\).

From the second equation, \(x = y + 1\). Substituting into the first: \(3(y + 1) + 2y = 12 \Rightarrow 5y + 3 = 12 \Rightarrow y = \dfrac{9}{5}\). Then \(x = y + 1 = \dfrac{14}{5}\). The solution is \(\left(\dfrac{14}{5}, \dfrac{9}{5}\right)\).

Intersection formula

The intersection of \(y = m_1 x + p_1\) and \(y = m_2 x + p_2\) (with \(m_1 \neq m_2\)) gives the point where both relations are simultaneously satisfied:

\[ x^* = \frac{p_2 - p_1}{m_1 - m_2}, \quad y^* = m_1 x^* + p_1. \]

First-Degree Inequalities

A first-degree inequality models a condition or a constraint — for example a maximum budget, a production capacity, or a break-even threshold. It is solved like an equation, with one crucial rule.

Rules for solving an inequality

Add or subtract the same number on both sides: the direction is unchanged.
Multiply or divide by a positive number: the direction is unchanged.
Multiply or divide by a negative number: the direction is reversed.

Warning

Forgetting to reverse the inequality sign when multiplying or dividing by a negative number is the most common mistake.

Graphically, the solution of an inequality corresponds to a region of the plane (a half-plane bounded by the line). For a system of inequalities the solution is the intersection of the admissible regions, which may be bounded or unbounded.

Worked example — solving an inequality

Solve \(-2x + 6 \geq 0\). Subtract \(6\): \(-2x \geq -6\). Divide by \(-2\) and reverse the sign: \(x \leq 3\). The solution is the half-line \(]-\infty, 3]\).

Synthesis exercise — logistics planning

A company offers two types of delivery:

express delivery (unit cost: \(40\) €);
standard delivery (unit cost: \(20\) €).

Each week the logistics manager plans for: a maximum budget of \(1{,}200\) €; no more than \(25\) express deliveries; and at least twice as many standard deliveries as express deliveries. Let \(x\) be the number of express deliveries and \(y\) the number of standard deliveries.

Questions and solution (click to expand)

1. Constraints. \[ 40x + 20y \leq 1200, \qquad x \leq 25, \qquad y \geq 2x, \qquad x \geq 0, \; y \geq 0. \] The budget constraint simplifies to \(2x + y \leq 60\).

2. Set of solutions. The admissible set is the intersection of these half-planes in the plane — the region satisfying all constraints simultaneously.

3. Is \((20, 45)\) admissible? Budget: \(40(20) + 20(45) = 800 + 900 = 1700 > 1200\). The budget constraint is violated, so the combination is not admissible.

4. If \(x = 10\), then \(y \geq 2(10) = 20\) and \(2(10) + y \leq 60 \Rightarrow y \leq 40\). So the number of standard deliveries lies between \(\mathbf{20}\) and \(\mathbf{40}\).

Practice exercises

Exercises (click for solutions)

Draw \(y = 2x + 3\) and \(y = -x + 5\) and find their intersection. Setting \(2x + 3 = -x + 5\) gives \(3x = 2\), \(x = \tfrac{2}{3}\), \(y = \tfrac{13}{3}\).
Solve \(\begin{cases} 3x + 2y = 12 \\ x - y = 1 \end{cases}\) → \(\left(\tfrac{14}{5}, \tfrac{9}{5}\right)\).
Sketch \(x \geq 0\), \(y \geq 0\), \(2x + y \leq 10\) and identify the feasible region (a triangle with vertices \((0,0)\), \((5,0)\), \((0,10)\)).

4 Worked Example

Worked example — comparing two phone plans

We model the monthly cost of each plan as a function of \(x\), the number of minutes of calls per month.

\[ C_A(x) = 10 + 0.05x, \qquad C_B(x) = 25. \]

Finding the break-even point. Set \(C_A(x) = C_B(x)\):

\[ 10 + 0.05x = 25 \;\Longrightarrow\; 0.05x = 15 \;\Longrightarrow\; x = 300 \text{ minutes.} \]

Interpretation.

At \(x = 300\) both plans cost €25.
For \(x < 300\) minutes, \(C_A(x) < C_B(x)\): Plan A is cheaper.
For \(x > 300\) minutes, \(C_A(x) > C_B(x)\): Plan B is cheaper.

A user who makes fewer than 300 minutes of calls per month should choose Plan A; a heavy user should choose Plan B.

5 Guided Practice

A company is deciding between two suppliers for a raw material. Supplier A charges a setup fee of €500 and €8 per unit. Supplier B charges no setup fee and €12 per unit.

Write the total cost function for each supplier.
Find the quantity at which both suppliers cost the same.
Which supplier is cheaper for an order of 100 units? Of 200 units?

Hint

Write \(C_A(q) = 500 + 8q\) and \(C_B(q) = 12q\). Set them equal to find the crossover quantity.

Show the solution

\(C_A(q) = 500 + 8q\), \(\quad C_B(q) = 12q\).
Equal when \(500 + 8q = 12q \Rightarrow 500 = 4q \Rightarrow q = 125\) units.
For \(q = 100\): \(C_A = 500 + 800 = 1300\), \(C_B = 1200\) → Supplier B is cheaper.
For \(q = 200\): \(C_A = 500 + 1600 = 2100\), \(C_B = 2400\) → Supplier A is cheaper.

6 Interactive Exploration

Use the sliders below to position two lines and observe how their intersection changes in real time.

viewof slope1 = Inputs.range([-5, 5], {
  value: 2,
  step: 0.5,
  label: "Slope of line 1 (m₁)"
})

viewof intercept1 = Inputs.range([-10, 10], {
  value: 1,
  step: 0.5,
  label: "Intercept of line 1 (p₁)"
})

viewof slope2 = Inputs.range([-5, 5], {
  value: -1,
  step: 0.5,
  label: "Slope of line 2 (m₂)"
})

viewof intercept2 = Inputs.range([-10, 10], {
  value: 7,
  step: 0.5,
  label: "Intercept of line 2 (p₂)"
})

parallel = Math.abs(slope1 - slope2) < 0.001
xStar = parallel ? NaN : (intercept2 - intercept1) / (slope1 - slope2)
yStar = parallel ? NaN : slope1 * xStar + intercept1

xRange = Array.from({length: 100}, (_, i) => -5 + i * 0.1)
lineData = xRange.flatMap(x => [
  {x, y: slope1 * x + intercept1, line: "Line 1"},
  {x, y: slope2 * x + intercept2, line: "Line 2"}
])

chart = Plot.plot({
  marks: [
    Plot.line(lineData, {x: "x", y: "y", stroke: "line"}),
    ...(parallel ? [] : [Plot.dot([{x: xStar, y: yStar}],
      {r: 5, fill: "red", title: d => `(${d.x.toFixed(2)}, ${d.y.toFixed(2)})`})])
  ],
  x: {domain: [-5, 5], label: "x"},
  y: {domain: [-10, 10], label: "y"},
  color: {legend: true}
})

html`
<div style="margin-bottom:1rem">${chart}</div>
<div class="metric-grid">
  <div class="metric-card">
    <span class="metric-label">Intersection x*</span>
    <strong>${parallel ? "No intersection (parallel lines)" : xStar.toFixed(2)}</strong>
  </div>
  <div class="metric-card">
    <span class="metric-label">Intersection y*</span>
    <strong>${parallel ? "—" : yStar.toFixed(2)}</strong>
  </div>
</div>
`

Investigation

Set \(m_1 = m_2\) (same slope). What happens to the intersection? What does this mean geometrically?
Model Plan A (\(m_1 = 0.05\), \(p_1 = 10\)) and Plan B (\(m_2 = 0\), \(p_2 = 25\)). Does the intersection match the worked example result?
Move both intercepts until the two lines coincide. What does this represent in a business context?

7 Applied Problem

A logistics manager must choose between two delivery contracts for the coming year. Contract A: fixed annual fee of €12,000 plus €2 per parcel delivered. Contract B: no fixed fee, €5 per parcel.

Write the annual cost function for each contract.
At what volume of parcels do both contracts cost the same?
The manager expects to deliver between 3,500 and 5,000 parcels. Which contract should she choose and why?
What constraint (inequality) describes the set of volumes for which Contract A is cheaper?

8 Common Errors

Common error — confusing slope and intercept

A student reads the line \(y = 3x - 2\) and states that the slope is \(-2\) and the \(y\)-intercept is \(3\).

The correct reading is: slope \(m = 3\) (coefficient of \(x\)) and \(y\)-intercept \(p = -2\) (constant term).

In \(y = mx + p\), the slope is always the coefficient of \(x\), not the standalone constant.

9 Check Your Understanding

Find the equation of the line through \((1, 4)\) and \((3, 10)\).
Two lines have the same slope but different \(y\)-intercepts. How many solutions does their system have?
A taxi costs €3.00 flagfall plus €1.80 per km. At what distance does it cost the same as a fixed-price ride of €21.00?
A student solves \(3x + 2y = 12\) and \(x - y = 1\) and gets \(x = 2\), \(y = 1\). Is this correct? Verify.
Solve \(-3x + 9 \geq 0\) and express the solution as an interval.

10 Summary

A line \(y = mx + p\) is fully described by its slope (rate of change) and its \(y\)-intercept (value when \(x = 0\)); the intersection of two lines is the point satisfying both equations simultaneously.
The intersection of two affine functions models equilibrium or crossover points in economics — the break-even quantity, the indifference point between two tariffs, or the market equilibrium price.
Inequalities model constraints (budgets, capacities, thresholds); the solution of a system of inequalities is a feasible region, which is the graphical tool for linear optimisation problems.

11 Independent Work

Solve systems of equations exercises 1–8 in the exercise bank.
Practise inequality problems 1–6.
Draw the feasible region for the logistics synthesis exercise.
Record errors in your error log.
Review glossary entries for: slope, intercept, system of equations, feasible region.

12 Continue

Continue to Module 4 →