Microeconomics, Linear Functions, and Exponential Models Study Guide

Course Logistics and Announcements

• Homework Schedule:

  • A homework assignment covering Section 1.4 is due tomorrow night.
  • This assignment is shorter and expected to take significantly less time than previous assignments.

• Quiz Schedule and Coverage:

  • The next quiz will take place on Wednesday.
  • The quiz covers Section 1.4 as well as Section 1.3 (since Section 1.3 was not assessed on the prior quiz).
  • Primary focus areas for preparation include understanding linear economic functions, how different functions interact, and their real-world interpretations.

• Previous Quiz Performance and Makeup Policy:

  • Average score: $6.5$ out of $8$ (approximately $81.25\%$, which is above $75\%$).
  • Grades remain unpublished due to several excused absences during the first full week requiring makeup quizzes.
  • Students with excused absences must coordinate makeup quizzes directly to allow grades to be finalized and published for the class.

Microeconomics and Business Functions (Section 1.4)

• Linear Cost Function Overview:

  • Basic linear cost equation model: $C(q) = \text{Fixed Cost} + (\text{Marginal Cost}) \times q$
  • Example cost function: $C(q) = 9876 + 0.5q$
  • $q$ represents the total quantity of goods produced.
  • Fixed Costs:
  • One-time expenditures required regardless of production volume (e.g., machinery, tooling, molds, factory facilities, patents).
  • Paid once and do not recur per unit produced.
  • Marginal Cost:
  • Represented mathematically by the slope of the linear cost function.
  • Represents the incremental cost incurred to produce one additional unit of a good (e.g., raw materials, direct labor).
  • Total cost $C(q)$ represents total money flowing out of the business.

• Revenue Function:

  • Formula: $\text{Revenue} = \text{Price} \times \text{Quantity}$
  • Mathematical expression: $R(q) = p \times q$
  • Represents total gross money incoming to the business from selling $q$ units at price $p$

• Profit Function:

  • Formula: $\text{Profit} = \text{Revenue} - \text{Cost}$
  • Mathematical expression: $P(q) = R(q) - C(q)$
  • Represents net financial gain (incoming revenue minus outgoing costs).
  • Businesses aim to maximize profit to allow for owner payout, reinvestment, and establishing business sustainability.

• Break-Even Point:

  • Occurs precisely when profit equals zero ($P(q) = 0$), or equivalently when revenue equals cost ($R(q) = C(q)$).
  • Crossing the break-even threshold is required for a business to generate positive net earnings and survive.

• Marginal Revenue and Marginal Profit:

  • Linear revenue and profit functions possess constant slopes known as marginal revenue and marginal profit.
  • Marginal Revenue: Additional revenue gained from selling one additional unit.
  • Marginal Profit: Additional profit gained from selling one additional unit.
  • Example: If producing and selling one additional good costs an extra $4\, \text{dollars}$ and generates an extra $50\, \text{dollars}$ in revenue, the marginal profit gained for that additional good is $46\, \text{dollars}$.

Supply and Demand Principles

• Economic Behavior of Supply and Demand:

  • Supply Curve Dynamics: As the market price of a good increases, suppliers are willing to produce and sell a larger quantity ($q$) because unit profitability rises.
  • Demand Curve Dynamics: As the market price of a good increases, consumer demand drops because purchasing the item becomes more expensive.
  • Real-World Consumer Example: High egg price inflation during COVID-19 led consumers to significantly reduce egg purchases and opt for alternative foods until prices dropped back down.

• Axis Conventions and Coordinate Setup:

  • Conceptually, price ($p$) acts as the independent variable determining the quantity ($q$) supplied or demanded.
  • Standard economic graphical representations place price ($p$) on the vertical y-axis and quantity ($q$) on the horizontal x-axis.

• Key Axis Intercepts and Extreme Points:

  • Supply Curve Y-Intercept ($p_0$):
  • Represents the minimum threshold price below which suppliers will produce zero units ($q = 0$).
  • Suppliers will not produce any goods unless the market price strictly exceeds $p_0$
  • Demand Curve Y-Intercept ($p_1$):
  • Represents the price point high enough that consumer quantity demanded drops to zero ($q = 0$).
  • Example: If egg prices reached $1000\, \text{dollars}$ per carton, zero units would be purchased by any consumer.
  • Demand Curve X-Intercept ($q_1$):
  • Represents the hypothetical quantity demanded if the good were completely free ($p = 0$).
  • Extreme Scenario Example: During COVID-19 economic disruptions, crude oil barrel spot prices dropped to negative values where energy companies paid entities to clear non-processed crude oil due to storage capacity limits.

Demand and Revenue Derivations

• Movie Theater Problem Setup:

  • Data Point 1: At price $p = 10\, \text{dollars}$, weekly ticket sales quantity $q = 50$
  • Data Point 2: At price $p = 8\, \text{dollars}$, weekly ticket sales quantity $q = 75$
  • Objective: Derive linear demand function $q(p)$ and total revenue function $R(p)$.

• Step-by-Step Demand Function Derivation:

  • Input variable $x = p$ (price in dollars), Output variable $y = q$ (quantity of tickets).
  • Points: $(10, 50)$ and $(8, 75)$
  • Slope calculation:     $m = \frac{q_2 - q_1}{p_2 - p_1} = \frac{75 - 50}{8 - 10} = \frac{25}{-2} = -12.5$
  • Point-Slope equation setup using $(10, 50)$:     $y - 50 = -\frac{25}{2}(x - 10)$
  • Algebraic simplification:     $y - 50 = -\frac{25}{2}x + 125$$y = -\frac{25}{2}x + 175$
  • Final Linear Demand Function:     $q(p) = -\frac{25}{2}p + 175$
  • Interpretation: The negative slope ($-\frac{25}{2}$) confirms that demand decreases as ticket prices rise.

• Step-by-Step Revenue Function Derivation:

  • Revenue formula expressed in terms of price input $p$:     $R(p) = q(p) \times p$
  • Substituting the linear demand expression into revenue:     $R(p) = \left(-\frac{25}{2}p + 175\right)p$
  • Distributing $p$:     $R(p) = -\frac{25}{2}p^2 + 175p$

Market Equilibrium Point

• Definition of Market Equilibrium:

  • The equilibrium point is the intersection point where quantity supplied equals quantity demanded ($q_{\text{supply}} = q_{\text{demand}}$).
  • In real markets, fluctuating market prices adjust over time until settling at this specific price and quantity.

• Equilibrium Problem Calculation:

  • Given Supply Curve: $q = 600p$
  • Given Demand Curve: $q = 33000 - 500p$ for $0 \le p \le 60$
  • Part A: Finding Equilibrium Price ($p$):
  • Set supply quantity equal to demand quantity:       $600p = 33000 - 500p$
  • Add $500p$ to both sides:       $1100p = 33000$
  • Solve for $p$:       $p = \frac{33000}{1100} = 30\, \text{units}$
  • Part B: Finding Equilibrium Quantity ($q$):
  • Substitute equilibrium price $p = 30$ into the supply function:       $q = 600(30) = 18000\, \text{units}$
  • Verification via demand function:       $q = 33000 - 500(30) = 33000 - 15000 = 18000\, \text{units}$

Comprehensive Business Application: Dog Biscuit Startup

• Problem Statement and Market Research Data:

  • Business Owner: Anna Marie
  • Initial market research point: At price $p = 5\, \text{dollars}$, sales quantity $q = 10100$ biscuits.
  • Price-demand sensitivity: Every price decrease of $0.50\, \text{dollars}$ increases sales quantity by $2772$ biscuits.

• Part A: Demand Function $q(p)$ Derivation:

  • Initial data point: $(5, 10100)$
  • Second data point calculated from rate of change:
  • New Price: $5 - 0.50 = 4.50\, \text{dollars}$
  • New Quantity: $10100 + 2772 = 12872\, \text{biscuits}$
  • Point 2: $(4.5, 12872)$
  • Slope calculation:     $m = \frac{\Delta q}{\Delta p} = \frac{2772}{-0.5} = -5544$
  • Point-Slope formulation using $(5, 10100)$:     $y - 10100 = -5544(x - 5)$$y - 10100 = -5544x + 27720$$y = -5544x + 37820$
  • Expressed as Demand Function $q(p)$:     $q(p) = -5544p + 37820$

• Part B: Total Revenue Function $R(p)$ Derivation:

  • General Revenue Equation: $R = p \times q$
  • Substitute demand expression $q(p)$ into revenue equation:     $R(p) = p(-5544p + 37820)$
  • Distribute price $p$:     $R(p) = -5544p^2 + 37820p$

Exponential Functions (Section 1.5)

• Consecutive Ratios vs. Absolute Differences:

  • Examining population growth data year-by-year shows that analyzing constant ratios provides meaningful patterns rather than simple linear differences.
  • Example Population Table Ratios across consecutive years:     $\frac{14.660}{14.235} \approx 1.030$$\frac{15.098}{14.660} \approx 1.030$$\frac{15.549}{15.098} \approx 1.030$
  • A consistent quotient confirms that the data possesses a constant growth factor.

• Multi-Year Ratio Analysis:

  • For non-consecutive years (e.g., a two-year gap between 2007 and 2009):     $\frac{f(2009)}{f(2007)} = \frac{15.098}{14.235} \approx 1.060$
  • Note that $1.060 = 1 + 2 \times 0.030$, illustrating how growth factors compound over multi-year spans.

• General Exponential Function Definition:

  • Mathematical form: $P(t) = P_0 a^t$
  • $P_0$: Initial quantity at time $t = 0$
  • $a$: Growth factor per unit time $t$
  • Relationship to percentage growth rate $r$: $a = 1 + r$
  • $r$ is the decimal equivalent of the percentage change.
  • Exponential Growth: Occurs when $a > 1$ (equivalent to $r > 0$).
  • Exponential Decay: Occurs when $0 < a < 1$ (equivalent to $r < 0$).

• Euler's Number ($e$):

  • Definition: $e \approx 2.71828$
  • Fundamental Calculus Property: The exponential function $e^x$ is its own derivative.
  • Formulates the base for the natural logarithm (denoted as $\ln(x)$), detailed further in Section 1.6.

Applied Exponential Practice Problems

• Adrenaline Level Decay and Growth:

  • Base Condition: Initial adrenaline level $P_0 = 15\, \text{mg}$. Formula structure: $P(t) = 15 a^t$
  • Scenario A: Increasing by $0.4$ per minute ($r = 0.4$):     $a = 1 + 0.4 = 1.4 \implies P(t) = 15(1.4)^t$
  • Scenario B: Decreasing by $0.4$ per minute ($r = -0.4$):     $a = 1 - 0.4 = 0.6 \implies P(t) = 15(0.6)^t$
  • Scenario C: Increasing by $3\%$ per minute ($r = 0.03$):     $a = 1 + 0.03 = 1.03 \implies P(t) = 15(1.03)^t$
  • Scenario D: Decreasing by $3\%$ per minute ($r = -0.03$):     $a = 1 - 0.03 = 0.97 \implies P(t) = 15(0.97)^t$

• Student Grades Model Analysis:

  • Problem Context: Evaluating six student grade tracking equations where time $t$ is measured in weeks after the first midterm exam.
  • Identification Task: Determine which equations represent Alex's grade, given that her grade has been continuously decreasing.
  • Solution:
  • Alex's grade is represented by formulas exhibiting exponential decay, which requires $a < 1$
  • Matching Equation Options:
    • Equation 3: $a = 0.9$
    • Equation 4: $a = 0.99$
    • Equation 6: $a = 0.89$
  • Underlying Principle: Bases strictly between $0$ and $1$ ($0 < a < 1$) continuously shrink the output value as exponent $t$ increases over time.