1.3 1.4 2.1 Comprehensive Study Notes: Linear Functions, Depreciation, Costs/Revenue/Profit, Demand & Supply, and Systems of Equations (Transcript-Based)

Functions: Definition and Notation

• A function f is a rule that assigns to each input value x exactly one output value y. Notation: $f(x)=y.$
• Domain: the set of all possible input x-values.
• Range: the set of all possible output y-values.
• Independent variable: x (the input).
• Dependent variable: y (the output).
• Example: For the linear function $y=5x+7$, the slope is the rate of change (m=5) and the y-intercept is the value when x=0 (b=7).
• In the form $y=f(x)$, f is the rule; x is the input, y is the output.

Linear Functions and Graphs

• A linear function has the form $y=mx+b$ where m is the slope and b is the y-intercept.
• Example: $y=3x+5$.
  • Value at x=0: $f(0)=5$ → point (0,5).
  • Value at x=1: $f(1)=8$ → point (1,8).
  • Value at x=10: $f(10)=35$ → point (10,35).
• If data points lie along a straight line, they can be modeled with a linear model.

Linear Depreciation (Example)

• Scenario: A machine costs $C_0=\$40{,}000$ and is depreciated linearly over 5 years by $\$5{,}000$ per year.
• Depreciation model (book value) as a function of year x: $B(x)=-5000\,x+40000.$
• After 3 years: $B(3)=-5000\cdot 3 + 40000 = 25000.$
• After 5 years (scrap value): $B(5)=-5000\cdot 5 + 40000 = 15000.$
• Slope = -5000, intercept = 40000.

Cost, Revenue, and Profit Functions

• Fixed cost b: a constant cost incurred regardless of production level.
• Variable cost per unit m: the cost that scales with production x.
• Total cost: $C(x)=m\,x+b.$
• Revenue: if selling price per unit is s, then $R(x)=s\,x.$ (Note: s is the price per unit.)
• Profit: $\Pi(x)=R(x)-C(x) = (s-m)x - b.$ If this is zero, you’re at break-even; if positive, profit; if negative, loss.
• Note: In production decisions, x is the quantity produced/sold, and the real-world interpretation depends on the units used for x and the scale of costs/revenue.

Break-even Analysis (General)

• Break-even point occurs when profit is zero: $\Pi(x)=0\Rightarrow (s-m)x=b\,.$
• If s>m, break-even quantity is $x_b=\dfrac{b}{s-m}.$
• If $s\le m$, there is no finite break-even point (either never profitable or always loss-making for positive x).
• At break-even, revenue equals cost: $R(x<em>b)=C(x</em>b).$

Unit Cost and Fixed Cost – Worked Example

• Given: cost function $C(x)=50x+6000$ and revenue function $R(x)=70x.$
  • Unit cost (variable cost per unit): $m=50.$
  • Fixed cost: $b=6000.$
  • Selling price per unit: $s=70.$
• Profit function: $P(x)=R(x)-C(x)=70x-(50x+6000)=20x-6000.$
• Break-even: solve $70x=50x+6000\Rightarrow 20x=6000\Rightarrow x=300.$
• At break-even, $P(300)=0.$

Break-even Calculation (Alternative Form)

• With break-even in the same setup, you can also compute using the profit form:
  $P(x)=(s-m)x-b.$
• Set to zero: $(s-m)x=b \Rightarrow x=\dfrac{b}{s-m}.$

Demand, Supply, and Equilibrium in Markets

• Demand function describes how quantity demanded responds to price; typically downward-sloping.
• Supply function describes how quantity supplied responds to price; typically upward-sloping.
• Equilibrium occurs where demand equals supply: $D(x)=S(x)$ at a common price p* and quantity x*.
• In graphs, the equilibrium is the intersection of the demand and supply curves.
• Note: In some worked examples, x is measured in thousands and p is in dollars.

Example: Solving a Demand–Supply System (2x2)

• Given a demand relation: $2x+7p=56,$ and a supply relation: $3x-11p=-45.$
• Solve the system for the equilibrium quantity x and price p.
• Method (elimination):
  • Multiply first equation by 3: $6x+21p=168.$
  • Multiply second equation by 2: $6x-22p=-90.$
  • Subtract to eliminate x: $43p=258\Rightarrow p=\frac{258}{43}=6.$
  • Back-substitute into 2x+7p=56: $2x+7\cdot 6=56\Rightarrow 2x+42=56\Rightarrow x=7.$
• Result: Equilibrium quantity $x^<em>=7$ (units in thousands), equilibrium price $p^</em>=\$6.$
• Note: The transcript shows p=16 due to a misprint; the correct arithmetic with the given system yields p=6.

Homework-Style Problem (Demand/Supply with 2 Linear Functions)

• Given: Demand function equation $4x+3p=59$ and Supply function equation $5x-6p=-14.$
• Find equilibrium price p* and quantity x*.
• Solve by equating expressions for x or by solving the system:
  • From demand: $4x=59-3p\Rightarrow x=\dfrac{59-3p}{4}.$
  • From supply: $5x-6p=-14\Rightarrow x=\dfrac{6p-14}{5}.$
  • Equate: $\dfrac{59-3p}{4}=\dfrac{6p-14}{5} \Rightarrow 5(59-3p)=4(6p-14)\Rightarrow 295-15p=24p-56.$
  • Solve: $351=39p\Rightarrow p^*=9.$
  • Then x* from either equation: $x^*=\dfrac{59-3(9)}{4}=\dfrac{32}{4}=8.$
• Result: Equilibrium price $p^<em>=\$9,$ equilibrium quantity $x^</em>=8$ (units as given by the problem, typically thousands).

Intersection of Lines (Two-Variable Case)

• For lines y = m1 x + b1 and y = m2 x + b2:
  • If m1 ≠ m2, intersection point is at $x<em>0=\dfrac{b</em>2-b<em>1}{m</em>1-m<em>2},\quad y</em>0=m<em>1 x</em>0+b_1.$
• If m1 = m2 and b1 ≠ b2, the lines are parallel (no solution).
• If m1 = m2 and b1 = b2, the lines coincide (infinitely many solutions).
• Example: Intersect y = x + 2 and y = -x + 7:
  • Solve x+2 = -x+7 ⇒ 2x = 5 ⇒ x0 = 2.5, and y0 = 2.5+2 = 4.5.
  • Intersection point: (2.5, 4.5).

Profit, Break-even, and Real-World With John’s Example

• Scenario: A product costs $C(x)=5x+12000$ and sells for $R(x)=11x.$
• Profit: $P(x)=R(x)-C(x)=11x-(5x+12000)=6x-12000.$
• Break-even: $6x-12000=0\Rightarrow x=2000.$
• Interpretation:
  • If production/sales x is 2000 units, profit is zero.
  • If x>2000, profit; if x<2000, loss.
• Example values from transcript (note: numbers may reflect unit scaling and should be checked in context):
  • C(100) = 5(100) + 12000 = 12500.
  • C(200) = 5(200) + 12000 = 13000.
  • R(200) = 11(200) = 2200.
  • R(500) = 11(500) = 5500.
• Key takeaway: The break-even quantity depends on fixed cost and the difference between selling price and variable cost.

Equilibrium Analysis in Markets (Summary)

• Equilibrium price and quantity occur where demand equals supply.
• If you write demand and supply as equations in p and x, solve the system to get (x, p).
• Interpretation of units matters: some examples use x in thousands and p in dollars; others use x in units.

3x3 Linear Systems and Planes (Geometric View)

• A 3×3 linear system Ax = b can be interpreted as the intersection of three planes in 3D space.
• A unique solution exists if det(A) ≠ 0 (the three planes intersect at a single point).
• No solution occurs when the planes do not intersect at a common point (inconsistent system).
• Infinitely many solutions occur when the planes coincide along a line or when the system is dependent.
• Example: Planes defined by three non-collinear points determine a unique plane. One example:
  • Through points (0,0,10), (0,5,0), and (10,0,0) the plane is $x+2y+z=10.$
• Gaussian elimination or matrix methods are standard tools to solve 3×3 systems.

3×3 Production Problem (Souvenirs A, B, C)

• Typical setup: three products A, B, C with time constraints on multiple machines (e.g., Machine I, Machine II, Machine III).
• Let x, y, z denote quantities of A, B, C produced.
• Constraints are linear (minutes per unit on each machine) with total available machine time.
• A worked example (illustrative):
  • 2x + 1y + 2z = 180 (Machine I capacity)
  • x + 3y + 2z = 300 (Machine II capacity)
  • 2x + 1y + 4z = 240 (Machine III capacity)
• Solving (Gaussian elimination) yields a specific production plan, e.g., (x, y, z) = (24, 72, 30).
• The general approach: set up Ax = b with A containing per-unit times, x the quantities, b the total times; solve for x.

Planes in 3D and Intersections (Extended View)

• A plane in 3D has equation ax + by + cz = d.
• Three non-collinear points determine a unique plane.
• A system of three planes (P1, P2, P3) can intersect at a single point (unique solution), along a line (infinitely many solutions), or be parallel/inconsistent (no solution).
• In matrix form, P1, P2, P3 can be represented by three equations, and their intersection is the solution to the corresponding 3×3 system, if it exists.

Final Problem: Two Linear Functions—Demand and Supply (Solving for Equilibrium)

• Given the two equations:
  • Demand: $4x+3p=59$
  • Supply: $5x-6p=-14$
• Solve for equilibrium price p* and quantity x*.
• Solve by equating expressions or by substitution:
  • From demand: $x=\dfrac{59-3p}{4}$
  • From supply: $x=\dfrac{6p-14}{5}$
  • Equate: $\dfrac{59-3p}{4}=\dfrac{6p-14}{5}$
  • Cross-multiply: $5(59-3p)=4(6p-14)$
  • Compute: $295-15p=24p-56\Rightarrow 351=39p\Rightarrow p^*=9.$
  • Solve for x: $x^</em>=\dfrac{59-3(9)}{4}=\dfrac{32}{4}=8.$
• Result: Equilibrium price $p^<em>=\$9$ and equilibrium quantity $x^</em>=8$ (units as specified; often x is in thousands).
• Interpretation: The demand function corresponds to p = (59-4x)/3 (as a function of x), which is downward-sloping in x; the supply function corresponds to p = (5x+14)/6, which is upward-sloping in x; their intersection gives the market equilibrium.