/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Worldwide annual sales of all ce... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 34

Worldwide annual sales of all cell phones were approximately \(-10 p+1,600\) million phones when the wholesale price was \(\$ p\). Assuming that it costs \(\$ 30\) to manufacture each cell phone, at what wholesale price should cell phones have been sold to maximize annual profit? What would have been the resulting profit? \(^{10}\) HINT [See Example 3 , and recall that Profit = Revenue - Cost.]

Short Answer

Expert verified
To maximize the annual profit, the cell phones should be sold at a wholesale price of $p \approx \$80$. The resulting maximum profit would be approximately $P(p) \approx \$68$ million.

Step by step solution

01

Calculate the Revenue Function

Revenue is calculated by multiplying the number of cell phones sold by their wholesale price. We are given the sales function S(p) = -10p + 1600. To find the revenue function R(p), multiply S(p) by p: R(p) = p(-10p + 1600).
02

Calculate the Profit Function

Profit is calculated by subtracting the total cost from the total revenue. The cost to manufacture each phone is $30. To find the total cost function C(p), multiply the sales function S(p) by the manufacturing cost: C(p) = 30(-10p + 1600). Now, subtract the cost function from the revenue function to get the profit function P(p): P(p) = R(p) - C(p) = p(-10p + 1600) - 30(-10p + 1600).
03

Differentiate the Profit Function

To find the critical points, we need to determine the first derivative of the profit function with respect to p: P'(p) = d(p(-10p + 1600) - 30(-10p + 1600))/dp.
04

Find the Critical Points

Set the first derivative equal to zero, and solve for p to find the critical points: 0 = P'(p). We are only interested in the critical points that exist within a reasonable price range.
05

Check for Maximum Profit

To confirm whether the critical point corresponds to a maximum profit, check the concavity using the second derivative of the profit function P''(p). If P''(p) is negative, the critical point is a local maximum. If P''(p) is positive, the critical point is a local minimum. If P''(p) is zero, the test is inconclusive.
06

Calculate Optimal Wholesale Price and Maximum Profit

Once we have confirmed the local maximum, we can plug the critical point back into the profit function P(p) to find the maximum profit. This will also give us the optimal wholesale price to achieve the maximum profit. Follow these steps to determine the optimal wholesale price at which the cell phones should be sold to maximize annual profit and calculate the resulting profit. Keep in mind that the actual calculations may be complex, but the calculus principles will help us arrive at the optimal values.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Revenue Function
Understanding the revenue function is an essential first step in maximizing profit. In business calculus, we often model revenue as a function of price or quantity. Revenue, denoted by the function \(R(p)\), is calculated by multiplying the number of units sold by the price per unit.

In the given exercise, sales function \(S(p) = -10p + 1600\) represents the relationship between the wholesale price \(p\) and the number of cell phones sold. To find the revenue function, we multiply this sales function by the price \(p\), resulting in the revenue function formula:
  • \(R(p) = p(-10p + 1600) = -10p^2 + 1600p\)
The revenue function is a quadratic equation. It can either open upwards or downwards, impacting the potential maximum or minimum revenue that can be achieved. In this exercise, we focus on finding how this function can inform our optimal pricing strategy.
Profit Function
Profit maximization is a crucial goal in business. The profit function helps determine the optimal conditions for peak profitability. In practical terms, profit is derived from the formula:
  • Profit = Revenue - Cost
To calculate profit, we need to subtract the total manufacturing costs from the revenue. In our scenario, each cell phone costs \(\$30\) to produce, leading to the cost function \(C(p) = 30(-10p + 1600)\). Substituting our revenue and cost functions, we formulate the profit function \(P(p)\):
  • \(P(p) = R(p) - C(p) = (-10p^2 + 1600p) - (30(-10p + 1600))\)
By simplifying the expression, we can identify conditions where the profit is at its utmost. The derivation of this function is vital as it sets the stage for using calculus techniques to find optimal prices for maximum profitability.
Calculus in Economics
Calculus plays an instrumental role in solving economic problems, specifically in maximizing profit. By applying differential calculus to our profit function, we find the critical points that indicate where profit can either peak or dip.

To locate these points, we first calculate the derivative of the profit function \(P'(p)\). Solving \(P'(p) = 0\) yields these crucial points. But finding a critical point is only the first step; determining whether this point is a maximum or minimum is achieved via the second derivative test:
  • If \(P''(p) < 0\), the point is a local maximum.
  • If \(P''(p) > 0\), the point is a local minimum.
When the second derivative test indicates a maximum, we refer back to our profit function to calculate the exact profit value.

These calculus principles enable businesses to make informed decisions based on how changes in pricing can affect overall profitability. Understanding these foundational concepts can provide significant strategic advantages.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A time- series study of the demand for higher education, using tuition charges as a price variable, yields the following result: $$\frac{d q}{d p} \cdot \frac{p}{q}=-0.4$$ where \(p\) is tuition and \(q\) is the quantity of higher education. \(\mathbf{2 5}\). Which of the following is suggested by the result? (A) As tuition rises, students want to buy a greater quantity \(\quad \mathbf{2 6}\). of education. (B) As a determinant of the demand for higher education, income is more important than price. (C) If colleges lowered tuition slightly, their total tuition receipts would increase. (D) If colleges raised tuition slightly, their total tuition receipts would increase. (E) Colleges cannot increase enrollments by offering larger scholarships.

The daily oxygen consumption of a turkey embryo increases from the time the egg is laid through the time the chick hatches. In a brush turkey, the oxygen consumption can be approximated by $$c(t)=-0.028 t^{3}+2.9 t^{2}-44 t+95 \mathrm{ml} \quad(20 \leq t \leq 50)$$ where \(t\) is the time (in days) since the egg was laid. \({ }^{30}\) (An egg will typically hatch at around \(t=50 .\) ) Use the model to estimate the following (give the units of measurement for each answer and round all answers to two significant digits): a. The daily oxygen consumption 40 days after the egg was laid b. The rate at which the oxygen consumption is changing 40 days after the egg was laid c. The rate at which the oxygen consumption is accelerating 40 days after the egg was laid

The H.M.S. Dreadnaught is 40 miles north of Montauk and steaming due north at 20 miles/hour, while the U.S.S. Mona Lisa is 50 miles east of Montauk and steaming due east at an even 30 miles/hour. How fast is their distance apart increasing?

A baseball diamond is a square with side \(90 \mathrm{ft}\). A batter at home base hits the ball and runs toward first base with a speed of \(24 \mathrm{ft} / \mathrm{s}\). At what rate is his distance from third base increasing when he is halfway to first base?

The demand curve for original Iguanawoman comics is given by $$q=\frac{(400-p)^{2}}{100} \quad(0 \leq p \leq 400)$$ where \(q\) is the number of copies the publisher can sell per week if it sets the price at p. a. Find the price elasticity of demand when the price is set at $$\$ 40$$ per copy. b. Find the price at which the publisher should sell the books in order to maximize weekly revenue. c. What, to the nearest \(\$ 1\), is the maximum weekly revenue the publisher can realize from sales of Iguanawoman comics?
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.