Price Elasticity of Demand - Business Calculus

A company sells two products, Alpha and Beta. At current prices, the price elasticity of demand for Alpha is $0.7$ , and for Beta is $2.4$ . If the company enacts a 5% price increase for both products, what is the most likely short-term impact on the total revenue generated by each product?
Answer: Revenue from Alpha will increase; revenue from Beta will decrease. When you encounter price elasticity problems, remember that elasticity measures how sensitive quantity demanded is to price changes, and this directly affects revenue outcomes. Price elasticity of demand compares the percentage change in quantity demanded to the percentage change in price. The key insight is that elasticity determines whether a price increase will increase or decrease total revenue. For Alpha (elasticity = 0.7): Since 0.7 < 1, demand is inelastic. When demand is inelastic, consumers aren't very responsive to price changes. A 5% price increase will cause quantity demanded to fall by less than 5% (approximately 3.5%). Since price rises more than quantity falls, total revenue increases. For Beta (elasticity = 2.4): Since 2.4 > 1, demand is elastic. Consumers are highly responsive to price changes. A 5% price increase will cause quantity demanded to fall by about 12%. Since quantity falls more than price rises, total revenue decreases. Therefore, Alpha's revenue increases while Beta's revenue decreases, making D correct. Option A is wrong because inelastic Alpha will see revenue increase, not decrease. Option B reverses the outcomes—it incorrectly suggests elastic demand increases revenue. Option C is wrong because elastic Beta will definitely lose revenue from a price increase. Study tip: Remember the elasticity rule: if elasticity < 1 (inelastic), price and revenue move in the same direction. If elasticity > 1 (elastic), price and revenue move in opposite directions. This relationship is crucial for pricing decisions in business calculus.
The relationship between price $p$ and quantity demanded $q$ for a certain commodity is given by the implicit equation $p^2 + 2q^2 = 1100$ . What is the price elasticity of demand when the price is $p = \$30$ ?
Answer: $4.50$ We need to find $E(p) = -\frac{p}{q} \cdot \frac{dq}{dp}$ . First, find the value of $q$ when $p=30$ : $30^2 + 2q^2 = 1100 \implies 900 + 2q^2 = 1100 \implies 2q^2 = 200 \implies q^2 = 100$ . Since quantity must be positive, $q=10$ . Next, find $\frac{dq}{dp}$ using implicit differentiation with respect to $p$ : $\frac{d}{dp}(p^2 + 2q^2) = \frac{d}{dp}(1100)$ $2p + 4q \frac{dq}{dp} = 0$ $\frac{dq}{dp} = -\frac{2p}{4q} = -\frac{p}{2q}$ . Now, evaluate $\frac{dq}{dp}$ at the point $(p,q) = (30,10)$ : $\frac{dq}{dp} = -\frac{30}{2(10)} = -\frac{30}{20} = -1.5$ . Finally, calculate the elasticity: $E(30) = -\frac{p}{q} \cdot \frac{dq}{dp} = -\frac{30}{10} \cdot (-1.5) = -3(-1.5) = 4.5$ .
The current price of a product is $p=\$40$ , and the price elasticity of demand is calculated to be $E(40) = 1.25$ . Based on this information, which action should the company take to increase total revenue?
Answer: Decrease the price, because demand is elastic, so the percentage increase in quantity will be greater than the percentage decrease in price. The price elasticity of demand is $E(40) = 1.25$ . Since $|E(p)| > 1$ , the demand is elastic at this price. For elastic demand, total revenue moves in the opposite direction of price. To increase total revenue, the company should decrease the price. A price decrease will lead to a proportionally larger increase in quantity demanded, thus increasing total revenue ( $R = p \cdot q$ ). Choice A correctly identifies the demand as elastic and prescribes the correct action with the correct reasoning.
The demand for a specialized microchip is given by the function $q(p) = 10000 e^{-0.05p}$ , where $p$ is the price in dollars. At what price is the demand unit elastic?
Answer: $p = \$20$ When you encounter elasticity problems in business calculus, you're measuring how responsive demand is to price changes. Unit elasticity occurs when the elasticity of demand equals -1, meaning a 1% price increase causes exactly a 1% decrease in quantity demanded. To find unit elasticity, you need the elasticity formula: $E(p) = \frac{p \cdot q'(p)}{q(p)}$ . First, find the derivative of the demand function $q(p) = 10000e^{-0.05p}$ . Using the chain rule: $q'(p) = 10000 \cdot (-0.05) \cdot e^{-0.05p} = -500e^{-0.05p}$ . Now substitute into the elasticity formula: $E(p) = \frac{p \cdot (-500e^{-0.05p})}{10000e^{-0.05p}} = \frac{-500p}{10000} = -0.05p$ For unit elasticity, set $E(p) = -1$ : $-0.05p = -1$ $p = 20$ Therefore, demand is unit elastic at $20. Choice A ($100) would give elasticity of -5, meaning demand is highly elastic. Choice B ($50) yields elasticity of -2.5, still elastic but not unit elastic. Choice D is incorrect because elasticity clearly varies with price—it's not constant. The elasticity function $E(p) = -0.05p$ shows elasticity changes linearly with price, and it equals -1 when $p = 20$ . Study tip: For exponential demand functions like $ae^{bp}$ , the elasticity is always $bp$ . This pattern can save you time—just set $bp = -1$ and solve for $p$ directly.
A tech company's subscription service has demand $p = 50e^{-0.02q}$ , where $p$ is monthly price and $q$ is thousands of subscribers. The marketing team claims that at their current subscriber level of 25,000, demand is unit elastic. If true, what should be their revenue-maximizing strategy?
Answer: Maintain current pricing since unit elasticity confirms they are already maximizing revenue at this point Unit elastic demand (elasticity = -1) occurs exactly at the revenue-maximizing point. When elasticity equals -1, the percentage change in price is exactly offset by the percentage change in quantity, keeping total revenue constant for small price changes. This is the theoretical maximum of the revenue function. Any price increase or decrease from this point will decrease total revenue. Choice B misunderstands that unit elasticity means revenue stays constant with price changes, not that margins improve. Choice C incorrectly suggests verifying the calculation rather than accepting the given information. Choice D misses that unit elasticity identifies the revenue maximum.
A streaming service has the demand function $q = 1000p^{-1.5}$ , where $q$ is in thousands of subscribers and $p$ is monthly price. The company is considering a price increase from $12 to $13. What is the expected impact on total revenue?
Answer: Revenue will decrease by approximately 4.2% because the price elasticity causes proportional quantity reduction For q = $1000p^{-1.5}$ , the elasticity is constant at -1.5 (the exponent). At p = 12: q = 1000 $(12)^{-1.5}$ ≈ 24.06, so R₁ = 12 × 24.06 ≈ 288.7. At p = 13: q = 1000 $(13)^{-1.5}$ ≈ 21.32, so R₂ = 13 × 21.32 ≈ 277.2. Revenue change = (277.2 - 288.7)/288.7 ≈ -4.0%. Since |elasticity| > 1, price increases decrease revenue. Choice A overestimates the revenue decrease. Choice C and D incorrectly suggest revenue increases, which cannot happen when demand is elastic.
The price elasticity of demand for a product at a price of $p_0$ is $E(p_0) = 0.8$ . At this price, the quantity demanded is $q_0 = 500$ units. Which of the following best describes the instantaneous rate of change of revenue with respect to price, $R'(p_0)$ ?
Answer: Revenue is increasing at a rate of $100 per dollar increase in price. The derivative of the revenue function, $R'(p)$ , is related to the price elasticity of demand, $E(p)$ , by the formula $R'(p) = q(1 - E(p))$ . We are given $q_0 = 500$ and $E(p_0) = 0.8$ . We can substitute these values into the formula to find the rate of change of revenue at this price point: $R'(p_0) = 500(1 - 0.8) = 500(0.2) = 100$ . Since $R'(p_0) = 100$ is positive, revenue is increasing at this price point. The value means that for a small increase in price, revenue increases by $100 times that change. Thus, revenue is increasing at a rate of $100 per dollar increase in price.
The demand function for a particular style of athletic shoe is given by $q(p) = 400(25-p^2)$ , where $q$ is the number of pairs sold per week and $p$ is the price in dollars, for $0 < p < 5$ . At what price is the demand for the shoes unit elastic?
Answer: $p = \frac{5\sqrt{3}}{3}$ Demand is unit elastic when the price elasticity of demand $E(p) = 1$ . First, find $q'(p) = 400(-2p) = -800p$ . The elasticity function is $E(p) = -\frac{p \cdot q'(p)}{q(p)} = -\frac{p(-800p)}{400(25-p^2)} = \frac{800p^2}{400(25-p^2)} = \frac{2p^2}{25-p^2}$ . Set $E(p) = 1$ : $\frac{2p^2}{25-p^2} = 1$ . This gives $2p^2 = 25-p^2$ , which simplifies to $3p^2 = 25$ . Solving for $p$ : $p^2 = \frac{25}{3}$ , so $p = \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}} = \frac{5\sqrt{3}}{3}$ .
The demand function for a brand of gourmet chocolate is given by $q = \frac{1800}{p+5} - 40$ , where $p$ is the price in dollars per bar and $q$ is the number of bars sold per week. At what price is the demand for the chocolate unit elastic?
Answer: $p = \$10$ The price elasticity of demand is given by the formula $E(p) = -\frac{p}{q} \cdot \frac{dq}{dp}$ . First, we find the derivative of the demand function: $q'(p) = \frac{d}{dp}(1800(p+5)^{-1} - 40) = -1800(p+5)^{-2} = -\frac{1800}{(p+5)^2}$ . Next, we set up the elasticity formula: $E(p) = -\frac{p}{\frac{1800}{p+5} - 40} \cdot \left(-\frac{1800}{(p+5)^2}\right) = \frac{1800p}{(\frac{1800 - 40(p+5)}{p+5})(p+5)^2} = \frac{1800p}{(1600 - 40p)(p+5)}$ . Demand is unit elastic when $E(p) = 1$ . So we set the expression equal to 1 and solve for $p$ : $1 = \frac{1800p}{(1600 - 40p)(p+5)}$ $(1600 - 40p)(p+5) = 1800p$ $1600p + 8000 - 40p^2 - 200p = 1800p$ $-40p^2 - 400p + 8000 = 0$ $p^2 + 10p - 200 = 0$ $(p+20)(p-10) = 0$ Since price must be positive, we have $p=10$ .
The total revenue $R(p)$ from selling a product at price $p$ is given by the function $R(p) = 300p^2 - 2p^3$ for the valid price range $0 \le p \le 150$ . For what range of prices is the demand for this product elastic?
Answer: $100 < p \le 150$ Demand is elastic when the price elasticity $E(p) > 1$ . A key relationship in business calculus is that demand is elastic if and only if revenue decreases as price increases, which means the derivative of the revenue function, $R'(p)$ , is negative. First, find the derivative of the revenue function: $R'(p) = 600p - 6p^2$ . Next, find the price range where $R'(p) < 0$ : $600p - 6p^2 < 0$ $6p(100 - p) < 0$ Since price $p$ must be positive, the term $6p$ is positive. For the entire expression to be negative, we must have $(100 - p) < 0$ , which implies $p > 100$ . Combining this with the given valid price range $0 \le p \le 150$ , the range for which demand is elastic is $100 < p \le 150$ .
A product has a linear demand function $q(p) = 240 - 2p$ , for the price range $0 \le p \le 120$ . At which price is the total revenue maximized?
Answer: $p = \$60$ Total revenue is maximized when the price elasticity of demand is unit elastic, i.e., $E(p)=1$ . For a linear demand function $q(p) = a - bp$ , revenue is maximized at the midpoint of its valid price range, which is $p = a/(2b)$ . Here, $a=240$ and $b=2$ . The revenue-maximizing price is $p = 240 / (2 \cdot 2) = 60$ . Alternatively, one can find the revenue function $R(p) = p \cdot q(p) = p(240 - 2p) = 240p - 2p^2$ . This is a downward-opening parabola, and its maximum occurs at the vertex. The p-coordinate of the vertex is $p = -\frac{B}{2A} = -\frac{240}{2(-2)} = 60$ .
The price elasticity of demand for a product is given by the function $E(p) = \frac{2p}{300-p}$ . At a price of $p=\$100$ , the demand is $q=8000$ units. Which of the following represents the demand function $q(p)$ ?
Answer: $q(p) = 0.2(300-p)^2$ The formula for elasticity is $E(p) = -\frac{p}{q}\frac{dq}{dp}$ . We are given $E(p)$ , so we can set up a differential equation: $\frac{2p}{300-p} = -\frac{p}{q}\frac{dq}{dp}$ . Assuming $p \neq 0$ , we can divide by $p$ : $\frac{2}{300-p} = -\frac{1}{q}\frac{dq}{dp}$ . This is a separable differential equation. Rearranging gives $\int \frac{2}{300-p} dp = \int -\frac{1}{q} dq$ . Integrating both sides yields: $-2 \ln(300-p) = -\ln(q) + C$ . (Assuming $p<300$ , so absolute values are not needed). Rearranging to solve for $q$ : $\ln(q) = 2 \ln(300-p) - C = \ln((300-p)^2) - C$ . Exponentiating both sides: $q = e^{\ln((300-p)^2) - C} = e^{-C}(300-p)^2$ . Let $k = e^{-C}$ . The demand function has the form $q(p) = k(300-p)^2$ . We use the point $(p=100, q=8000)$ to find the constant $k$ : $8000 = k(300-100)^2 = k(200)^2 = 40000k$ . Solving for $k$ : $k = \frac{8000}{40000} = \frac{1}{5} = 0.2$ . Thus, the demand function is $q(p) = 0.2(300-p)^2$ .
A movie theater determines that its daily demand function for tickets is $q(p) = 1600 - p^2$ for a price $p$ . The theater is currently charging $p = \$20$ per ticket. To increase total revenue from ticket sales, what action should the theater take based on the price elasticity of demand?
Answer: Increase the price, because demand is inelastic at $p=\$20$ . When you encounter price elasticity questions, you need to determine whether demand is elastic or inelastic at the current price, then apply the revenue rule: if demand is elastic, lower prices increase revenue; if demand is inelastic, higher prices increase revenue. Given the demand function $q(p) = 1600 - p^2$ , you first calculate the price elasticity of demand using $E_d = \frac{dq/dp \cdot p}{q}$ . Taking the derivative: $\frac{dq}{dp} = -2p$ . At $p = 20$ , the quantity is $q(20) = 1600 - 400 = 1200$ . The elasticity becomes: $E_d = \frac{-2(20) \cdot 20}{1200} = \frac{-800}{1200} = -\frac{2}{3}$ Since $|E_d| = \frac{2}{3} < 1$ , demand is inelastic at this price point. When demand is inelastic, consumers are relatively unresponsive to price changes, so the theater can increase prices to boost total revenue. Answer D correctly identifies both that demand is inelastic and that the theater should increase prices. Answer A incorrectly states demand is elastic when it's actually inelastic. Answer B correctly identifies the inelastic nature but wrongly suggests decreasing prices, which would reduce revenue when demand is inelastic. Answer C suggests the right action (increase prices) but incorrectly claims demand is elastic. Remember this key relationship: elastic demand means lower prices increase revenue, while inelastic demand means higher prices increase revenue. Always calculate the elasticity first, then apply the appropriate pricing strategy.
A pharmaceutical company finds that the demand for their medication follows $q = \frac{1000}{p^{0.8}}$ . If the company wants to increase total revenue and current elasticity of demand is -0.8, what pricing strategy should they pursue and why?
Answer: Increase price because demand is inelastic and higher prices will increase revenue despite lower quantity When |elasticity| < 1, demand is inelastic. With inelastic demand, a price increase leads to a proportionally smaller decrease in quantity, resulting in higher total revenue. The elasticity of -0.8 means a 1% price increase causes only a 0.8% quantity decrease, yielding net revenue gain. Choice B incorrectly classifies -0.8 as elastic demand. Choice C wrongly suggests revenue is maximized when elasticity = -0.8 (revenue maximization occurs when elasticity = -1). Choice D makes an irrelevant connection between the elasticity value and the demand function exponent.
An analyst determines that the price elasticity of demand for a monthly software subscription is $0.75$ at the current price. The company is considering increasing the price. Which of the following is the most accurate conclusion that can be drawn from this information alone?
Answer: Total revenue will increase, but the effect on profit cannot be determined without cost information. The price elasticity of demand is $E = 0.75$ . Since $E < 1$ , the demand is inelastic. For inelastic demand, an increase in price leads to an increase in total revenue. This is because the percentage decrease in quantity demanded will be smaller than the percentage increase in price. So, we can conclude that total revenue will increase. However, profit is calculated as Total Revenue - Total Cost. While revenue will increase, a price increase will also cause the quantity sold to decrease. The change in total cost depends on how cost varies with quantity (i.e., the marginal cost). Without information on the cost structure, we cannot determine whether the increase in revenue will outweigh the change in total production costs. Therefore, the effect on profit is uncertain.
The weekly demand for a smart home device is modeled by the function $q = 200e^{-0.05p}$ , where $q$ is the number of units sold and $p$ is the price in dollars. The current price is $p = \$30$ . To increase total revenue, how should the company adjust the price?
Answer: Lower the price, because the demand is elastic at this price point. First, calculate the price elasticity of demand $E(p)$ . The derivative of the demand function is $q'(p) = 200e^{-0.05p} \cdot (-0.05) = -10e^{-0.05p}$ . The elasticity formula is $E(p) = -\frac{p}{q} \cdot q'(p)$ . $E(p) = -\frac{p}{200e^{-0.05p}} \cdot (-10e^{-0.05p}) = \frac{10p}{200} = 0.05p$ . Now, evaluate the elasticity at the current price $p = 30$ : $E(30) = 0.05(30) = 1.5$ . Since $E(30) = 1.5 > 1$ , demand is elastic. When demand is elastic, a price decrease leads to an increase in total revenue. Therefore, the company should lower the price.
The demand function for a product is given by $q = \frac{k}{p}$ , where $q$ is the quantity demanded, $p$ is the price, and $k$ is a positive constant. Which of the following correctly describes the price elasticity of demand for this product?
Answer: The elasticity is 1 at all prices. To find the price elasticity of demand, we use the formula $E(p) = -\frac{p}{q} \cdot \frac{dq}{dp}$ . The demand function is $q = kp^{-1}$ . First, we find the derivative with respect to price: $\frac{dq}{dp} = -1 \cdot kp^{-2} = -\frac{k}{p^2}$ . Now, we substitute $q$ and $\frac{dq}{dp}$ into the elasticity formula: $E(p) = -\frac{p}{k/p} \cdot \left(-\frac{k}{p^2}\right)$ . Simplifying the expression: $E(p) = -\frac{p^2}{k} \cdot \left(-\frac{k}{p^2}\right) = \frac{p^2k}{kp^2} = 1$ . The elasticity is equal to 1 for all values of $p$ and $k$ . This special type of demand curve is a rectangular hyperbola, for which total revenue ( $R = p \cdot q = p \cdot (k/p) = k$ ) is constant. When revenue is constant, demand is always unit elastic.
The demand for a product is modeled by the function $q(p) = \frac{5000}{p^{1.5}}$ . Which statement accurately describes the effect of a price change on total revenue?
Answer: A price decrease will always increase revenue, because demand is always elastic. For a demand function of the form $q(p) = \frac{k}{p^n} = kp^{-n}$ , the price elasticity of demand is constant and equal to the exponent $n$ . Here, $n=1.5$ . We can verify this: $q'(p) = -1.5 k p^{-2.5}$ . $E(p) = -\frac{p \cdot q'(p)}{q(p)} = -\frac{p(-1.5kp^{-2.5})}{kp^{-1.5}} = 1.5 \frac{kp^{-1.5}}{kp^{-1.5}} = 1.5$ . Since the elasticity is $E(p) = 1.5$ for all prices $p$ , which is greater than 1, demand is always elastic. When demand is elastic, price and total revenue move in opposite directions. Thus, a price decrease will always cause an increase in total revenue.
The demand function for a particular model of tablet computer is given by $q = \frac{5000}{p^2}$ , where $p$ is the price per tablet. If the current price is $p = \$20$ , what is the approximate percentage change in demand that would result from a 1% increase in the price?
Answer: A decrease of approximately 2% The relationship between elasticity and percentage change is $E \approx -\frac{\%\Delta q}{\%\Delta p}$ . We first need to find the price elasticity of demand $E(p)$ . The demand function is $q = 5000p^{-2}$ . The derivative is $q'(p) = -10000p^{-3} = -\frac{10000}{p^3}$ . The elasticity is $E(p) = -\frac{p}{q} \cdot q'(p) = -\frac{p}{5000/p^2} \cdot \left(-\frac{10000}{p^3}\right) = \frac{p^3}{5000} \cdot \frac{10000}{p^3} = 2$ . The elasticity of demand is constant at $E(p)=2$ . Now, we use the approximation formula with $E=2$ and $\%\Delta p = +1\%$ : $2 \approx -\frac{\%\Delta q}{1\%}$ . Solving for $\%\Delta q$ , we get $\%\Delta q \approx -2\%$ . Therefore, a 1% price increase will lead to an approximate 2% decrease in demand.
The relationship between the price $p$ (in dollars) and the quantity demanded $q$ (in thousands of units) for a product is given by the equation $p^2 + 2q^2 = 1100$ . What is the price elasticity of demand when the price is $p=\$30$ ?
Answer: 4.5 This problem requires three steps. First, find the quantity $q$ when $p=30$ . Substitute into the equation: $(30)^2 + 2q^2 = 1100 \implies 900 + 2q^2 = 1100 \implies 2q^2 = 200 \implies q^2 = 100 \implies q=10$ (since quantity must be positive). Second, find the derivative $\frac{dq}{dp}$ using implicit differentiation: $2p + 4q\frac{dq}{dp} = 0 \implies \frac{dq}{dp} = -\frac{2p}{4q} = -\frac{p}{2q}$ . Third, calculate the price elasticity of demand, $E(p) = -\frac{p}{q} \frac{dq}{dp}$ . $E(p) = -\frac{p}{q} \left(-\frac{p}{2q}\right) = \frac{p^2}{2q^2}$ . Finally, substitute the values $p=30$ and $q=10$ : $E(30) = \frac{(30)^2}{2(10)^2} = \frac{900}{2(100)} = \frac{900}{200} = 4.5$ .

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