3️⃣ Curve Analysis in Economics and Management
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.
- 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.
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 \(x = 0\)).
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.
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.
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 \(\left(\dfrac{14}{5}, \dfrac{9}{5}\right)\).
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.
- 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.
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.
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
- Draw \(y = 2x + 3\) and \(y = -x + 5\) and find their intersection. Setting \(2x + 3 = -x + 5\) gives \(3x = 2\), \(x = \tfrac23\), \(y = \tfrac{13}{3}\).
- 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 \((0,0)\), \((5,0)\), \((0,10)\)).