Building Linear Models - Finite Mathematics

1. An employee earns $20 per hour for the first 40 hours worked in a week. For any hours worked beyond 40, the employee earns an overtime rate of 1.5 times the standard rate. Which linear model represents the employee's total weekly pay, W, for working h hours, where h > 40?
   Answer: $W(h) = 30h - 400$ For $h > 40$, the total pay W is the sum of the pay for the first 40 hours and the pay for the overtime hours. Pay for the first 40 hours is $40 \times \$20 = \$800$. The overtime rate is $1.5 \times \$20 = \$30$ per hour. The number of overtime hours is $h - 40$. The overtime pay is $30(h - 40)$. So, the total pay is $W(h) = 800 + 30(h - 40) = 800 + 30h - 1200 = 30h - 400$.
2. The relationship between temperature in degrees Fahrenheit (F) and degrees Celsius (C) is linear. The freezing point of water is 0°C and 32°F, and the boiling point of water is 100°C and 212°F. Which of the following models correctly expresses F as a function of C?
   Answer: $F = \frac{9}{5}C + 32$ When you encounter a temperature conversion problem, you're working with linear functions where you need to find the relationship between two variables using given data points. To find the correct function F = f(C), start with the two known points: (0°C, 32°F) and (100°C, 212°F). Since this is a linear relationship, you can find the slope using $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{212 - 32}{100 - 0} = \frac{180}{100} = \frac{9}{5}$ Using the point-slope form with point (0, 32): $F - 32 = \frac{9}{5}(C - 0)$, which simplifies to $F = \frac{9}{5}C + 32$. You can verify this works: when C = 0, F = 32; when C = 100, F = 212. Looking at the wrong answers: Choice A gives you $C = \frac{9}{5}F + 32$, which solves for C in terms of F (the reverse of what's asked). Choice B has $F = \frac{5}{9}C + 32$, which uses the reciprocal slope—this would give you F = 32 + 20 = 52°F when C = 100°C, clearly wrong since water boils at 212°F. Choice C gives $F = \frac{9}{5}(C - 32)$, which incorrectly subtracts 32 from C before multiplying; this would give negative temperatures for normal Celsius values. Study tip: In linear conversion problems, always identify your two data points first, calculate the slope carefully, and double-check by plugging in the original values. Watch out for reciprocal slopes and sign errors—these are the most common traps.
3. A small business manufactures artisanal chairs. Its fixed monthly costs include $2,000 for rent and $500 for equipment leasing. The variable cost to produce one chair is $15. If each chair sells for $40, which linear model represents the monthly profit P as a function of the number of chairs x produced and sold?
   Answer: $P(x) = 25x - 2500$ Profit P is Revenue R minus Cost C. The total fixed cost is $2000 + 500 = $2500. The variable cost is $15x$. So the total cost function is $C(x) = 15x + 2500$. The revenue function is $R(x) = 40x$. Therefore, the profit function is $P(x) = R(x) - C(x) = 40x - (15x + 2500) = 40x - 15x - 2500 = 25x - 2500$.
4. A water tank is being filled at a constant rate. After 15 minutes, the tank contains 240 gallons. After 35 minutes, it contains 320 gallons. If the tank was not empty when filling began, what was the initial amount of water in the tank?
   Answer: 180 gallons, found by extending the linear relationship back to time zero The filling rate is $(320 - 240)/(35 - 15) = 80/20 = 4$ gallons per minute. Using point-slope form with (15, 240): $y - 240 = 4(t - 15)$, so $y = 4t + 180$. The initial amount (at $t = 0$) is 180 gallons. Choice A assumes the tank started empty, Choice C incorrectly subtracts 15 minutes worth of water (240 - 60 = 180, but this logic is flawed), and Choice D gives an incorrect y-intercept calculation.
5. Based on the passage, which of the following equations correctly models the quantity demanded, q, as a linear function of the price, p?
   Answer: $q = -5p + 1100$ The demand data gives two points (p, q): (120, 500) and (100, 600). The slope of the demand function is $m = \frac{q_2 - q_1}{p_2 - p_1} = \frac{600 - 500}{100 - 120} = \frac{100}{-20} = -5$. The model is $q = -5p + b$. Using the point (100, 600) to find the intercept b: $600 = -5(100) + b \Rightarrow 600 = -500 + b \Rightarrow b = 1100$. Thus, the demand model is $q = -5p + 1100$.
6. A town's population was 12,500 in the year 2010. By 2015, the population had grown to 14,000. Assuming the population growth is linear, which model represents the population P as a function of t, the number of years after 2010?
   Answer: $P(t) = 300t + 12500$ When you encounter a linear population growth problem, you're working with a straight line relationship where the population changes by the same amount each year. You need to find the slope (rate of change) and y-intercept (starting value) to build your linear function. Start by identifying your two data points: (0, 12,500) for 2010 and (5, 14,000) for 2015. Since t represents years after 2010, year 2010 is t = 0 and 2015 is t = 5. Calculate the slope: $m = \frac{14,000 - 12,500}{5 - 0} = \frac{1,500}{5} = 300$ people per year. Using the point-slope form with our starting point (0, 12,500), the function becomes $P(t) = 300t + 12,500$. You can verify this: when t = 0, P = 12,500 ✓; when t = 5, P = 300(5) + 12,500 = 14,000 ✓. Option A gives $P(t) = 300t - 590,500$, which has the correct slope but a massive negative y-intercept that would predict a negative population in 2010. Option B shows $P(t) = 1,500t + 12,500$, using the total population change (1,500) as the slope instead of the annual rate (300). Option D reverses the slope and y-intercept, creating $P(t) = 12,500t + 300$, which would predict unrealistic exponential-like growth. Remember: in linear growth problems, always calculate the rate per unit time as your slope, and use your initial value as the y-intercept. Double-check by substituting both given data points back into your equation.
7. A company has fixed costs of $10,000 per month and a variable cost of $30 per unit. The product sells for $50 per unit. The company aims to achieve a target monthly profit of $5,000. Which equation correctly models the number of units, x, that must be sold to reach this target profit?
   Answer: $50x - (30x + 10000) = 5000$ The profit P is calculated as total revenue R(x) minus total cost C(x). Here, $R(x) = 50x$ and $C(x) = 30x + 10000$. The profit function is $P(x) = 50x - (30x + 10000)$. To find the number of units needed for a target profit of $5,000, we set the profit function equal to the target: $50x - (30x + 10000) = 5000$.
8. Let x be the number of units of Product X and y be the number of units of Product Y produced daily. Based on the passage, which linear inequality correctly models the constraint on available machine time?
   Answer: $15x + 25y \leq 480$ The machine time for x units of Product X is $15x$ minutes. The machine time for y units of Product Y is $25y$ minutes. The total machine time is $15x + 25y$. The total available time is 8 hours, which must be converted to minutes to match the units of the coefficients: $8 \text{ hours} \times 60 \text{ minutes/hour} = 480$ minutes. The total time used must be less than or equal to the available time, so the constraint is $15x + 25y \leq 480$.
9. A nutritionist is creating a food supplement from two sources, A and B. Let a represent the number of grams of source A and b represent the number of grams of source B. A dietary requirement specifies that the amount of source A must be at least two-thirds the amount of source B. Which of the following linear inequalities, written in a standard form, correctly models this constraint?
   Answer: $3a - 2b \geq 0$ The statement 'the amount of source A must be at least two-thirds the amount of source B' translates to the inequality $a \geq \frac{2}{3}b$. To convert this to a standard form with integer coefficients and no fractions, we can multiply both sides by 3: $3a \geq 2b$. Finally, to move all variables to one side, we subtract $2b$ from both sides, which gives $3a - 2b \geq 0$.
10. A fitness tracker app models the relationship between weekly exercise minutes $m$ and calories burned $c$ using the equation $c = 6.5m + 120$. The constant term represents calories burned through normal daily activities. If a user wants to burn at least 800 calories per week through this combination, but can only exercise for a maximum of 90 minutes per week, what is the range of exercise minutes that meets their goal?
    Answer: The goal cannot be met within the exercise time constraint of 90 minutes per week To burn at least 800 calories: $6.5m + 120 \geq 800$, so $6.5m \geq 680$, which means $m \geq 104.6$ minutes. Since the user can only exercise for a maximum of 90 minutes per week, and they need at least 104.6 minutes to reach their goal, it's impossible to meet the 800-calorie target within the time constraint. The other choices incorrectly assume the goal is achievable within 90 minutes.
11. A manufacturing company's monthly production cost $C$ (in thousands of dollars) is modeled by $C = 12 + 0.8x$, where $x$ is the number of units produced (in hundreds). If the company's budget allows for a maximum monthly cost of $28,000, and they need to produce at least 500 units to meet demand, which statement correctly describes their production constraints?
    Answer: They can produce between 5 and 20 hundred units while staying within budget and meeting demand First, convert units: 500 units = 5 hundred units (minimum), and $28,000 = 28 thousand dollars (maximum cost). The cost constraint gives us: $12 + 0.8x \leq 28$, so $0.8x \leq 16$, thus $x \leq 20$ hundred units. Combined with the minimum requirement of $x \geq 5$ hundred units, they can produce between 5 and 20 hundred units. Choice B uses the wrong units (individual units vs. hundreds), Choice C ignores the budget constraint, and Choice D is incorrect since the constraints are feasible.
12. A coffee blender is creating a 100-pound batch of a new blend by mixing two types of coffee beans. Bean A costs $5.00 per pound and Bean B costs $8.00 per pound. Let x represent the number of pounds of Bean A used. Which of the following linear models represents the total cost, C, of the 100-pound batch as a function of x?
    Answer: $C(x) = 800 - 3.00x$ Let x be the pounds of Bean A and y be the pounds of Bean B. The total cost is $C = 5x + 8y$. The total weight is 100 pounds, so $x + y = 100$. To express the cost as a function of x, we solve for y: $y = 100 - x$. Substitute this into the cost equation: $C(x) = 5x + 8(100 - x) = 5x + 800 - 8x = 800 - 3x$.
13. A piece of industrial equipment is purchased and depreciates linearly over time. Two years after purchase, its value is $45,000. Five years after purchase, its value is $30,000. Which linear model represents the value V of the equipment t years after its purchase?
    Answer: $V(t) = -5000t + 55000$ The problem gives two points (t, V): (2, 45000) and (5, 30000). First, find the slope (rate of depreciation): $m = \frac{V_2 - V_1}{t_2 - t_1} = \frac{30000 - 45000}{5 - 2} = \frac{-15000}{3} = -5000$. The model is $V(t) = -5000t + b$. Use one point to find the y-intercept b (initial value): $45000 = -5000(2) + b \Rightarrow 45000 = -10000 + b \Rightarrow b = 55000$. Thus, the model is $V(t) = -5000t + 55000$.
14. A research study tracks the relationship between study time and test scores. The linear model $S = 65 + 2.5t$ represents the relationship, where $S$ is the test score and $t$ is study time in hours. However, the model is only valid for study times between 2 and 20 hours. What is the range of possible test scores predicted by this model?
    Answer: From 70 to 115 points, accounting for the minimum and maximum valid study times Since the model is only valid for $2 \leq t \leq 20$, we find the score range by evaluating at the endpoints. At $t = 2$: $S = 65 + 2.5(2) = 70$. At $t = 20$: $S = 65 + 2.5(20) = 115$. So the range is 70 to 115 points. Choice A uses the y-intercept (which is outside the valid domain), Choice C uses incorrect endpoint calculations, and Choice D extends beyond the valid maximum.

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