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Chapter 4: Problem 18

The demand equation for a product is \(p=45-0.01 q\) Write the revenue as a function of \(q\) and find the quantity that maximizes revenue. What price corresponds to this quantity? What is the total revenue at this price?

Short Answer

Expert verified
The quantity that maximizes revenue is 2250 units, the corresponding price is $22.5, and the total revenue is $50,625.

Step by step solution

01

Write the Revenue Function

The revenue \(R\) is calculated as the product of the price \(p\) and the quantity \(q\). Given the demand equation \(p = 45 - 0.01q\), we can express revenue as a function of \(q\):\[ R(q) = p \cdot q = (45 - 0.01q) \cdot q = 45q - 0.01q^2 \]
02

Find the Quantity that Maximizes Revenue

To find the quantity \(q\) that maximizes revenue, we need to take the derivative of the revenue function \(R(q) = 45q - 0.01q^2\) with respect to \(q\), and set it equal to zero.\[ \frac{dR}{dq} = 45 - 0.02q = 0 \]Solving for \(q\), we get:\[ 0.02q = 45 \]\[ q = \frac{45}{0.02} = 2250 \]
03

Find the Price Corresponding to the Optimal Quantity

Substitute \(q = 2250\) into the demand equation to find the corresponding price \(p\):\[ p = 45 - 0.01 \times 2250 = 45 - 22.5 = 22.5 \]
04

Calculate the Total Revenue at Optimal Quantity and Price

Using the revenue function \(R(q) = 45q - 0.01q^2\), substitute \(q = 2250\) to calculate total revenue:\[ R(2250) = 45 \times 2250 - 0.01 \times 2250^2 = 101250 - 50625 = 50625 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Demand Equation
To understand how the demand equation impacts revenue maximization, let's first explore what it represents. A demand equation shows the relationship between the price of a product and the quantity demanded by consumers. In this context, we're using the simple linear equation:

\( p = 45 - 0.01q \)

This equation suggests that as more units of product are sold, the lower the price becomes. It's common for higher quantities to push down the price consumers are willing to pay. Demand equations are essential for businesses to understand market behavior and play a significant role in determining pricing strategies. They help forecast how changes in price affect demand, and consequently, potential revenue.
Revenue Function
The revenue function is at the heart of any business looking to maximize its profit. Revenue is calculated by multiplying the price per unit by the quantity of units sold. In our problem, we express revenue as a function of quantity using the demand equation:

\( R(q) = p \cdot q = (45 - 0.01q) \cdot q = 45q - 0.01q^2 \)

This equation gives us the total revenue possible for every level of demand, accounting for the relationship between price and quantity as described by the demand equation. Understanding and calculating the revenue function allows businesses to predict the income generated from different sales volumes, which is crucial for strategic decision-making.
Derivative
Calculating the derivative of a function is a powerful method in mathematics used to find the rate of change. For revenue maximization, we need to take the derivative of the revenue function with respect to quantity because it helps identify the maximum point of the function. For the given revenue function

\( R(q) = 45q - 0.01q^2 \)

we find the derivative by applying basic calculus rules:

\( \frac{dR}{dq} = 45 - 0.02q \)

Setting this derivative equal to zero allows us to find the quantity that maximizes revenue. By solving \( 45 - 0.02q = 0 \), we determine \( q = 2250 \). This process of finding the derivative supports businesses in making informed decisions on optimal production levels to maximize revenue.
Optimized Pricing
Optimized pricing is crucial for maximizing revenue while considering market demand. Once the optimal quantity is found (in this case, \( q = 2250 \) units), we can use the demand equation to find the corresponding price. Substituting the optimal quantity into the demand equation:

\( p = 45 - 0.01 \times 2250 = 22.5 \)

shows that at 2250 units, the market price should be \( 22.5 \) per unit. In practice, optimized pricing not only considers internal factors but also external market conditions. It's an essential tool for businesses to ensure competitiveness, customer satisfaction, and profitability. Calculating the total revenue at this price, using the revenue function, gives practical insights into financial expectations.

\( R(2250) = 50625 \)

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Most popular questions from this chapter

The average cost per item to produce \(q\) items is given by $$ a(q)=0.01 q^{2}-0.6 q+13, \text { for } q>0 $$ (a) What is the total cost, \(C(q),\) of producing \(q\) goods? (b) What is the minimum marginal cost? What is the practical interpretation of this result? (c) At what production level is the average cost a minimum? What is the lowest average cost? (d) Compute the marginal cost at \(q=30 .\) How does this relate to your answer to part (c)? Explain this relationship both analytically and in words.

What are the units of elasticity if: (a) Price \(p\) is in dollars and quantity \(q\) is in tons? (b) Price \(p\) is in yen and quantity \(q\) is in liters? (c) What can you conclude in general?

A demand function is \(p=400-2 q,\) where \(q\) is the quantity of the good sold for price \(\$ p\) (a) Find an expression for the total revenue, \(R\), in terms of \(q\) (b) Differentiate \(R\) with respect to \(q\) to find the marginal revenue, \(M R,\) in terms of \(q .\) Calculate the marginal revenue when \(q=10\) (c) Calculate the change in total revenue when production increases from \(q=10\) to \(q=11\) units. Confirm that a one-unit increase in \(q\) gives a reasonable approximation to the exact value of \(M R\) obtained in part (b).

Show analytically that if marginal cost is less than average cost, then the derivative of average cost with respect to quantity satisfies \(a^{\prime}(q)<0\)

A warehouse selling cement has to decide how often and in what quantities to reorder. It is cheaper, on average, to place large orders, because this reduces the ordering cost per unit. On the other hand, larger orders mean higher storage costs. The warehouse always reorders cement in the same quantity, \(q .\) The total weekly cost, \(C,\) of ordering and storage is given by \(C=\frac{a}{q}+b q, \quad\) where \(a, b\) are positive constants. (a) Which of the terms, \(a / q\) and \(b q,\) represents the ordering cost and which represents the storage cost? (b) What value of \(q\) gives the minimum total cost?
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