Problem 54
Question
By the demand curve for a given commodity, we mean the set of all points \((p, q)\) in the \(p q\) plane where \(q\) is the number of units of the product that can be sold at price \(p .\) Use the differential approximation to estimate the demand \(q(p)\) for a commodity at a given price \(p\). The demand curve for a commodity is given by $$ \frac{p^{2} q}{10}+5 p \sqrt{q}=39000 $$ when \(p\) is measured in dollars. Approximately how many units of the commodity can be sold at \(\$ 1.80 ?\)
Step-by-Step Solution
Verified Answer
Approximately 110 units can be sold at $1.80.
1Step 1: Differentiate the Given Equation
The given demand curve function is \( \frac{p^2 q}{10} + 5p \sqrt{q} = 39000 \). We'll differentiate this equation with respect to \( p \) to find \( \frac{dq}{dp} \). Each term must be differentiated separately.
2Step 2: Apply Chain Rule
Differentiate the first term, \( \frac{p^2 q}{10} \), with respect to \( p \) using the product rule: \( \frac{d}{dp}\left( \frac{p^2 q}{10} \right) = \frac{2pq}{10} + \frac{p^2}{10}\frac{dq}{dp} \).
3Step 3: Differentiate the Second Term
Differentiate the second term, \( 5p \sqrt{q} \), with respect to \( p \): \( \frac{d}{dp}\left( 5p \sqrt{q} \right) = 5 \sqrt{q} + \frac{5p}{2\sqrt{q}}\frac{dq}{dp} \).
4Step 4: Set Up the Equation
Combine the differentiated terms to form: \( \frac{2pq}{10} + \frac{p^2}{10}\frac{dq}{dp} + 5\sqrt{q} + \frac{5p}{2\sqrt{q}}\frac{dq}{dp} = 0 \).
5Step 5: Solve for \( \frac{dq}{dp} \)
Isolate \( \frac{dq}{dp} \): \( \frac{dq}{dp} \left( \frac{p^2}{10} + \frac{5p}{2\sqrt{q}} \right) = -\left( \frac{2pq}{10} + 5\sqrt{q} \right) \). Thus, \( \frac{dq}{dp} = \frac{-\left( \frac{2pq}{10} + 5\sqrt{q} \right)}{\left( \frac{p^2}{10} + \frac{5p}{2\sqrt{q}} \right)} \).
6Step 6: Estimate \( q \) at \( p = 1.80 \)
Substitute \( p = 1.80 \) into the original equation to find \( q \). You get \( \frac{(1.80)^2 q}{10} + 5(1.80) \sqrt{q} = 39000 \). Solve this equation to find the approximate value of \( q \).
7Step 7: Find \( \frac{dq}{dp} \) at \( p = 1.80 \)
Use the expression found in Step 5 to compute \( \frac{dq}{dp} \) at \( p = 1.80 \) and determined \( q \) from the previous step. This will provide the approximate change in \( q \) with respect to a small change in \( p \).
8Step 8: Calculate the Approximate Quantity Sold
Using the linear approximation formula \( q(p) \approx q(p_0) + \frac{dq}{dp}(p - p_0) \), estimate \( q \) for \( p = 1.80 \). Set \( p_0 = 2.00 \) or any nearby known price and substitute the values to find the approximate quantity sold.
Key Concepts
Demand CurveChain RuleProduct RuleDifferential Approximation
Demand Curve
The demand curve is an essential concept in economics. It represents the relationship between the price of a commodity and the quantity demanded by consumers. This typically forms a downward sloping line on a chart when plotted, indicating that as prices drop, the quantity demanded usually increases, and vice versa.
In the given exercise, the demand curve is represented by the equation:\[\frac{p^{2} q}{10} + 5 p \sqrt{q} = 39000\]
In this equation:
In the given exercise, the demand curve is represented by the equation:\[\frac{p^{2} q}{10} + 5 p \sqrt{q} = 39000\]
In this equation:
- \(p\) is the price of the good measured in dollars.
- \(q\) is the number of units bought at price \(p\).
Chain Rule
The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. If you have a function composed of two or more functions, the chain rule allows you to find the derivative efficiently. In our exercise, the chain rule helps simplify the differentiation of complex terms.
For instance, consider the term \(5p \sqrt{q}\). Using the chain rule, we focus on simplifying the differentiation by treating \(\sqrt{q}\) as a separate function. The differentiation expression is:\[\frac{d}{dp}(5p \sqrt{q}) = 5 \sqrt{q} + \frac{5p}{2\sqrt{q}}\frac{dq}{dp}\]
Understanding the chain rule is crucial because it unlocks the ability to differentiate terms that are products or composed of nested functions children.
For instance, consider the term \(5p \sqrt{q}\). Using the chain rule, we focus on simplifying the differentiation by treating \(\sqrt{q}\) as a separate function. The differentiation expression is:\[\frac{d}{dp}(5p \sqrt{q}) = 5 \sqrt{q} + \frac{5p}{2\sqrt{q}}\frac{dq}{dp}\]
Understanding the chain rule is crucial because it unlocks the ability to differentiate terms that are products or composed of nested functions children.
Product Rule
The product rule is another important concept in calculus used to differentiate the product of two functions. When you have two functions multiplied together, the product rule tells you how to find the derivative of the product.
The product rule states that for two functions \(u(p)\) and \(v(p)\), the derivative \(\frac{d}{dp}(uv)\) is given by:\[\frac{d}{dp}(uv) = u'v + uv'\]
In our exercise, when differentiating the term \(\frac{p^2 q}{10}\), we apply the product rule. Here, \(u = \frac{p^2}{10}\) and \(v = q\), and thus their derivative using the product rule becomes:\[\frac{d}{dp}\left(\frac{p^2q}{10}\right) = \frac{2pq}{10} + \frac{p^2}{10}\frac{dq}{dp}\]
Using the product rule simplifies complex derivatives and is indispensable in dealing with terms composed of multiple components.
The product rule states that for two functions \(u(p)\) and \(v(p)\), the derivative \(\frac{d}{dp}(uv)\) is given by:\[\frac{d}{dp}(uv) = u'v + uv'\]
In our exercise, when differentiating the term \(\frac{p^2 q}{10}\), we apply the product rule. Here, \(u = \frac{p^2}{10}\) and \(v = q\), and thus their derivative using the product rule becomes:\[\frac{d}{dp}\left(\frac{p^2q}{10}\right) = \frac{2pq}{10} + \frac{p^2}{10}\frac{dq}{dp}\]
Using the product rule simplifies complex derivatives and is indispensable in dealing with terms composed of multiple components.
Differential Approximation
Differential approximation simplifies the calculation of small changes in functions by using linear approximations. It involves estimating the change in function value with respect to a change in its input, a key concept when exact values are hard to calculate but estimates are needed.
In this exercise, differential approximation helps in estimating the number of units of the commodity \(q\) sold at a specific price \(p = 1.80\). The linear approximation formula used is:\[q(p) \approx q(p_0) + \frac{dq}{dp}(p - p_0)\]
This formula is particularly valuable when determining approximations around a known price point \(p_0\), allowing us to predict how the demand will change with a small variation in price. Differential approximation is helpful in making quick predictions and informed decisions based on mathematical estimates rather than lengthy computations.
In this exercise, differential approximation helps in estimating the number of units of the commodity \(q\) sold at a specific price \(p = 1.80\). The linear approximation formula used is:\[q(p) \approx q(p_0) + \frac{dq}{dp}(p - p_0)\]
This formula is particularly valuable when determining approximations around a known price point \(p_0\), allowing us to predict how the demand will change with a small variation in price. Differential approximation is helpful in making quick predictions and informed decisions based on mathematical estimates rather than lengthy computations.
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