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Chapter 3: Problem 35

For the price function $$p(x)=10-\sqrt{3 x} $$where \(p\) is in dollars and \(x\) is the number sold in hundreds \((0 \leq x \leq 30),\) use \(R(12)\) and \(M R(12)\) to approximate the revenue when 1220 items are sold, and interpret the marginal revenue value. How much will the price need to be changed to have this level of sales?

Short Answer

Expert verified
Set the price at $3.95 per item to sell 1220 units; marginal revenue is $8.26.

Step by step solution

01

Define Revenue Function

The revenue function can be calculated by multiplying the price per item by the number of items sold. The price function given is \(p(x) = 10 - \sqrt{3x}\). Therefore, the revenue function, \(R(x)\), is \(R(x) = x \cdot p(x) = x(10 - \sqrt{3x})\). Here, \(x\) represents the number of items sold in hundreds.
02

Calculate Revenue for 1,220 Items

We need to determine the revenue for \(x = 12.2\) (as 1,220 items sold implies 1220/100 hundreds of items). Substitute \(x = 12.2\) into \(R(x)\): \[ R(12.2) = 12.2 (10 - \sqrt{3 \times 12.2}) = 12.2 (10 - \sqrt{36.6}) = 12.2 (10 - 6.05) \]. This equals \(12.2 \times 3.95 = 48.19\).
03

Compute Marginal Revenue Function

To find the marginal revenue, first differentiate the revenue function \(R(x)\) with respect to \(x\). The derivative \(R'(x)\) represents the marginal revenue, \(M R(x)\). Differentiating \(R(x) = 10x - x\sqrt{3x}\), we apply the product rule and simple differentiation: \(R'(x) = 10 - \frac{3x}{2\sqrt{3x}}\). Simplifying, \(R'(x) = 10 - \frac{\sqrt{x}}{2}\).
04

Calculate Marginal Revenue at 12.2

Insert \(x = 12.2\) into the marginal revenue formula: \(M R(12.2) = 10 - \frac{\sqrt{12.2}}{2}\). Calculate \(\sqrt{12.2} \approx 3.49\); \(M R(12.2) = 10 - \frac{3.49}{2} = 10 - 1.745 = 8.255\).
05

Interpret Marginal Revenue

The marginal revenue \(M R(12.2) = 8.255\) indicates that each additional 100 items sold generates approximately $8.26 in revenue before taking cost into account. This reflects the rate of change of revenue with respect to selling more items.
06

Determine Price Change for 1,220 Units

To maintain 1,220 units sales, we use the price function \(p(x)\) for \(x=12.2\). We substitute \(x = 12.2\) into \(p(x) = 10 - \sqrt{3 \times 12.2} = 10 - 6.05 = 3.95\). The price needs to be set at approximately \$3.95 per item for 1,220 units.

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Key Concepts

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

Price Function
The price function is a key element when calculating revenue. In this exercise, the price function is expressed as \(p(x) = 10 - \sqrt{3x}\). Here, \(p(x)\) denotes the price in dollars of each item based on the sales volume, \(x\). This function shows the relationship between the number of items sold, measured in hundreds, and the price. We notice that as \(x\) increases, \(p(x)\) decreases, indicating a decrease in price as more items are sold. This is typical in pricing strategies where bulk purchases lead to discounts.
  • The formula \(p(x) = 10 - \sqrt{3x}\) captures how price inversely relates to sales volume.
  • This understanding helps businesses strategize pricing according to demand.
Marginal Revenue
Marginal revenue (MR) is crucial for understanding the additional gain from selling one more unit. It is derived from the revenue function, representing the additional revenue gained per additional hundred items, in this context. From our exercise, we found the marginal revenue when \(x = 12.2\) using our derived formula, \(M R(x) = 10 - \frac{\sqrt{x}}{2}\).
  • Plugging in \(x = 12.2\), we calculate \(M R(12.2) \approx 8.255\).
  • This result shows that selling an additional 100 items around this level of sales increases revenue by approximately \$8.26.
  • Understanding MR helps businesses decide whether increasing production will lead to increased earnings.
Marginal revenue provides insight into pricing and output decisions, illustrating how revenue responds to changes in sales volume.
Derivative
The derivative is a vital mathematical concept used to determine the rate of change. In this scenario, it helps us find the marginal revenue from the revenue function. The process of differentiation finds the slope of the revenue curve at any point, indicating how revenue changes as more items are sold.
  • We differentiate \(R(x) = 10x - x \sqrt{3x}\) to get \(R'(x) = 10 - \frac{\sqrt{x}}{2}\).
  • The derivative simplifies into a straightforward formula to analyze MR.
  • Knowing derivatives allows businesses to predict future revenue trends.
The use of derivatives transforms our understanding of static revenue into dynamic predictions, vital for decision-making.
Revenue Function
The revenue function encapsulates how much a business earns based on the price and quantity sold. In this problem, the revenue function is formulated as \(R(x) = x (10 - \sqrt{3x})\), where \(x\) is the number of hundreds of items sold. This expression combines both the price function and the sale volume to determine total revenue.
  • We can evaluate \(R(x)\) to find the specific revenue for any sales level, such as \(R(12.2) = 48.19\) dollars for 1220 items sold.
  • This function provides a clear framework for anticipating total earnings.
  • Understanding the revenue function allows businesses to tailor strategies to maximize income.
With the revenue function, businesses hold a powerful tool to predict and achieve desired financial outcomes by manipulating price and sales strategies.

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Most popular questions from this chapter

In each equation, \(x\) and \(y\) are functions of \(t\) Differentiate with respect to \(t\) to find a relation between \(d x / d t\) and \(d y / d t\). $$ x^{3}-x y=y^{3} $$

A company's cost function is $$ C(x)=x^{2}+2 x+4 $$ dollars, where \(x\) is the number of units. a. Enter the cost function in \(y_{1}\) on a graphing calculator. b. Define \(y_{2}\) to be the marginal cost function by defining \(y_{2}\) to be the derivative of \(y_{1}\) (using NDERIV). c. Define \(y_{3}\) to be the company's average cost function, $$A C(x)=\frac{C(x)}{x}$$ by defining \(y_{3}=\frac{y_{1}}{x}\). d. Turn off the function \(y_{1}\) so that it will not be graphed, but graph the marginal cost function \(y_{2}\) and the average cost function \(y_{3}\) on the window [0,10] by \([0,10] .\) Observe that the marginal cost function pierces the average cost function at its minimum point (use TRACE to see which curve is which function). e. To see that the final sentence of part (d) is true in general, change the coefficients in the cost function \(C(x),\) or change the cost function to a cubic or some other function [so that \(C(x) / x\) has a minimum]. Again turn off the cost function and graph the other two to see that the marginal cost function pierces the average cost function at its minimum.

The number \(x\) of printer cartridges that a store will sell per week and their price \(p\) (in dollars) are related by the equation \(x^{2}=4500-5 p^{2}\). If the price is falling at the rate of \(\$ 1\) per week, find how the sales will change if the current price is \(\$ 20 .\)

For the cost function$$C(x)=x+10 \sqrt{x+9}$$ where \(C\) is in dollars and \(x\) is the number produced in hundreds, use \(C(7)\) and \(M C(7)\) to approximate the cost of producing 620 items. Give an interpretation of the marginal cost value.

True or False: At a critical number the function must be defined.
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