/*! 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 2 Find the maximum value of the fu... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 2

Find the maximum value of the function \(f(x)=12 x-x^{2}\), and give the value of \(x\) where this maximum occurs.

Short Answer

Expert verified
The maximum value of the function is 36, occurring at x=6.

Step by step solution

01

- Identify the function

The given function is: \[f(x) = 12x - x^2\]
02

- Find the first derivative

Differentiate the function with respect to x: \[f'(x) = \frac{d}{dx}(12x - x^2) = 12 - 2x\]
03

- Set the first derivative to zero

To find the critical points, set the first derivative equal to zero and solve for x: \[12 - 2x = 0\]\[2x = 12\]\[x = 6\]
04

- Verify the critical point is a maximum

To confirm that x=6 is a maximum, we can use the second derivative test. Compute the second derivative of the function: \[f''(x) = \frac{d}{dx}(12 - 2x) = -2\]Since the second derivative is negative, the critical point at x=6 is a local maximum.
05

- Find the maximum value

Substitute x=6 back into the original function to find the maximum value of f(x): \[f(6) = 12(6) - (6)^2 = 72 - 36 = 36\]Hence, the maximum value of the function is 36.

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.

Critical Points
Critical points are important in finding the maximum or minimum values of a function. A critical point occurs where the first derivative of a function equals zero or does not exist. These points are where the function changes direction, allowing us to identify peaks and troughs.
For instance, to find the critical points of the function given in the exercise:
  • We first determine the first derivative: \(f'(x)=12-2x\)
  • Set the first derivative equal to zero to solve for x: \(12-2x=0\)
  • Solve for x: \(x=6\)
So, the critical point of the function is at \(x=6\). This is where we will check if a maximum or minimum value occurs.
First Derivative
The first derivative of a function provides valuable information about the function's slope and can help determine critical points. Calculating the first derivative involves differentiating the function with respect to its independent variable.
In our given function \(f(x)=12x-x^2\), we find the first derivative by performing the following steps:
  • Differentiating the function to get \(f'(x)=12-2x\)
The first derivative \(f'(x)=12-2x\) represents the rate of change of the function. By setting it to zero, \(12-2x=0\), we solve for the critical point at \(x=6\). This process helps us to locate where the function's slope is zero, indicating a potential maximum or minimum point.
Second Derivative Test
After finding the critical points using the first derivative, we apply the second derivative test to determine whether those points are maxima, minima, or points of inflection. The second derivative test examines the concavity of the function at critical points.
In this exercise, after identifying our critical point \(x=6\), we calculate the second derivative of the function \(f(x)=12x-x^2\):
  • First find the first derivative: \(f'(x)=12-2x\)
  • Differentiating \(f'(x)\) gives the second derivative: \(f''(x)=-2\)
Since \(f''(x)=-2\) is negative, it indicates concavity downwards. This means the function reaches a maximum at \(x=6\). So, we confirm that \(x=6\) is a maximum point.
Quadratic Function
Quadratic functions are polynomial functions of degree two, represented by the general form \(f(x)=ax^2+bx+c\). They graph as parabolas which can open upwards or downwards depending on the sign of the leading coefficient (a).
The function in the exercise \(f(x)=12x-x^2\) is of the form \(ax^2+bx+c\) with \(a=-1\), \(b=12\), and \(c=0\):
  • Since \(a=-1\) (negative), the parabola opens downwards, implying the vertex is a maximum point.
  • The vertex/critical point is where we earlier found \(x=6\).
By substituting \(x=6\) back into the original function, we find the maximum value of \(f(x)\). So, we compute: \(f(6)=12(6)-(6)^2=72-36=36\). Therefore, the function's maximum value is 36 at \(x=6\).

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

Sketch the following curves, indicating all relative extreme points and inflection points. $$ y=\frac{1}{3} x^{3}-x^{2}-3 x+5 $$

A one-product firm estimates that its daily total cost function (in suitable units) is \(C(x)=x^{3}-6 x^{2}+13 x+15\) and its total revenue function is \(R(x)=28 x\). Find the value of \(x\) that maximizes the daily profit.

The canopy height (in meters) of the tropical bunch-grass elephant millet \(t\) days after mowing (for \(t \geq 32\) is \(f(t)=-3.14+.142 t-.0016 t^{2}+\) \(.0000079 t^{3}-.0000000133 t^{4} .\) (Source: Crop Science.) (a) Graph \(f(t)\) in the window \([32,250]\) by \([-1.2,4.5]\). (b) How tall was the canopy after 100 days? (c) When was the canopy 2 meters high? (d) How fast was the canopy growing after 80 days? (e) When was the canopy growing at the rate of \(.02 \mathrm{me}-\) ters per day? (f) Approximately when was the canopy growing slowest? (g) Approximately when was the canopy growing fastest?

Coffee consumption in the United States is greater on a per capita basis than anywhere else in the world. However, due to price fluctuations of coffee beans and worries over the health effects of caffeine, coffee consumption has varied considerably over the years. According to data published in The Wall Street Journal, the number of cups \(f(x)\) consumed daily per adult in year \(x\) (with 1955 corresponding to \(x=0)\) is given by the mathematical model $$ f(x)=2.77+0.0848 x-0.00832 x^{2}+0.000144 x^{3} $$ (a) Graph \(y=f(x)\) to show daily coffee consumption from 1955 through 1994 . (b) Use \(f^{\prime}(x)\) to determine the year in which coffee consumption was least during this period. What was the daily coffee consumption at that time? (c) Use \(f^{\prime}(x)\) to determine the year in which coffee consumption was greatest during this period. What was the daily coffee consumption at that time? (d) Use \(f^{\prime \prime}(x)\) to determine the year in which coffee consumption was decreasing at the greatest rate.

Find two positive numbers, \(x\) and \(y\), whose sum is 100 and whose product is as large as possible.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Logic and Functions

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.