/*! 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 27 Find the marginal profit for pro... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the marginal profit for producing units. (The profit is measured in dollars.) $$ P=-2 x^{2}+72 x-145 $$

Short Answer

Expert verified
The marginal profit for producing x units is given by \( P'(x) = -4x + 72 \)

Step by step solution

01

Understand the problem

The profit as a function of the number of produced units is given as \( P=-2 x^{2}+72 x-145 \). Marginal profit is the derivative of the profit function with respect to x, which tells us how the profit changes as the output (x) increases or decreases. Our objective is to find this marginal profit, i.e., to differentiate \( P \) with respect to \( x \).
02

Apply the power rule of derivatives

Taking the derivative of the profit function \(P\) with respect to \(x\) (denoted by \(P'(x)\) or \(dP/dx\)): Applying the power rule which states that if \(f(x) = ax^n\), then \(f'(x) = nax^{n-1}\), we find: \[ P'(x) = -2 \cdot 2x^{2-1} + 72 \cdot 1x^{1-1} - 0 \cdot 145 = -4x + 72 \]. The term -0 \cdot 145 drops out as the derivative of a constant is zero.
03

Evaluate the derivative

We derived that the marginal profit function \(dP/dx\) or \(P'(x)\) equals -4x + 72. This equation tells us the rate of change of the profit as we produce more or less units. The coefficient -4 indicates that for each additional unit we produce, the profit decreases by 4 dollars, and the 72 indicates the profit at production level zero.

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 91Ó°ÊÓ!

Key Concepts

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

Derivative of a Function
In calculus, the derivative of a function represents how a function changes as its input changes. It is the fundamental concept used to determine the rate at which one quantity changes with respect to another. For a function like the profit function provided,\[ P = -2x^2 + 72x - 145, \] the derivative indicates how the profit changes as we alter the number of units produced, represented by \( x \). Let's say the function is \( f(x) \), with a derivative denoted by \( f'(x) \) or \( \frac{df}{dx} \). This derivative gives us a new function that describes the slope of the tangent line to \( f(x) \) at any point. In the context of the profit function, calculating the derivative helps us understand the behavior of profit as production quantity changes. Analyzing the derivative is crucial for making informed business decisions.
  • Understanding derivatives allows businesses to maximize profit.
  • They provide insights into optimal production levels.
  • Help forecast the effects of changes in production.
Rate of Change in Profit
The rate of change in profit is essentially the marginal profit. This concept tells you how much profit increases or decreases with each additional unit produced. Once you have the derivative of the profit function, \( P'(x) \), it gives you direct insight into this rate. From our example, the derivative was calculated as \( P'(x) = -4x + 72 \). This expression is crucial because:
  • -4 signifies that for every additional unit produced, profit decreases by 4 dollars. This is the marginal effect on profit.
  • 72 represents the potential initial contribution to profit when no units are produced, often referred to as the intercept.
Understanding the rate of change in profit allows businesses to align their production strategies with profitability goals. They can predict when increasing production will no longer be beneficial and make adjustments accordingly. By evaluating \( P'(x) \) at specific values of \( x \), companies can forecast profit trends based on production scenarios. It acts as a preview to management, helping foresee which output level leads to maximum or minimum profit.
Power Rule of Derivatives
The power rule of derivatives is a fundamental rule in calculus used to find the derivative of polynomial functions. It simplifies the differentiation process, especially for functions involving powers of \( x \). Essentially, if you have a function in the form \( f(x) = ax^n \), its derivative is \( f'(x) = nax^{n-1} \). For our problem, we apply the power rule to the profit function:- The first term: \( -2x^2 \) becomes \( -4x \) as per the power rule: \( -2 \cdot 2x^{2-1} = -4x \).- The second term: \( 72x \) becomes \( 72 \) because \( 72 \cdot 1x^{0} = 72 \). - The constant term \(-145\) disappears, as constants have a derivative of zero.Using the power rule helps us smoothly transition from a complex profit function to a simpler expression, which tells us the marginal profit. This rule is essential in calculus, mainly because it streamlines the differentiation process of functions, making it quicker and less error-prone. Memorizing and practicing this rule will significantly empower your problem-solving skills in various fields, including economics and business studies.

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

Find \(f^{\prime}(x)\) $$ f(x)=\left(3 x^{2}-5 x\right)\left(x^{2}+2\right) $$

Find the marginal profit for producing units. (The profit is measured in dollars.) $$ P=-0.25 x^{2}+2000 x-1,250,000 $$

Health The temperature \(T\) (in degrees Fahrenheit) of a person during an illness can be modeled by the equation \(T=-0.0375 t^{2}+0.3 t+100.4,\) where \(t\) is time in hours since the person started to show signs of a fever. (a) Use a graphing utility to graph the function. Be sure to choose an appropriate window. (b) Do the slopes of the tangent lines appear to be positive or negative? What does this tell you? (c) Evaluate the function for \(t=0,4,8,\) and \(12 .\) (d) Find \(d T / d t\) and explain its meaning in this situation. (e) Evaluate \(d T / d t\) for \(t=0,4,8,\) and \(12 .\)

(a) sketch the graphs of \(f\) and \(g,(b)\) find \(f^{\prime}(1)\) and \(g^{\prime}(1),(c)\) sketch the tangent line to each graph when \(x=1,\) and \((d)\) explain the relationship between \(f^{\prime}\) and \(g^{\prime}\). $$ \begin{array}{l}{f(x)=x^{2}} \\ {g(x)=3 x^{2}}\end{array} $$

Use the table to answer the questions below. $$ \begin{array}{|cc|cc|}\hline \text { Quantity } & {} & {} & {} \\ {\text { produced }} & {} & {\text { Total }} & {\text { Marginal }} \\ {\text { and sold }} & {\text { Price }} & {(T R)} & {(M R)} \\ {(Q)} & {(p)} & {} & {(M R)} \\ \hline 0 & {160} & {0} & {-} \\ {2} & {140} & {280} & {130} \\ {4} & {120} & {480} & {90} \\ {6} & {100} & {600} & {50} \\ {8} & {80} & {640} & {10} \\ {10} & {60} & {600} & {-30} \\ \hline\end{array} $$ (a) Use the regression feature of a graphing utility to find a quadratic model that relates the total revenue \((T R)\) to the quantity produced and sold \((Q) .\) (b) Using derivatives, find a model for marginal revenue from the model you found in part (a). (c) Calculate the marginal revenue for all values of \(Q\) using your model in part (b), and compare these values with the actual values given. How good is your model?

See all solutions

Recommended explanations on Math Textbooks

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.