/*! 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 374 What rectangle of maximum area c... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 374

What rectangle of maximum area can be inscribed in a circle of radius \(\mathrm{r}\) ?

Short Answer

Expert verified
The dimensions of the rectangle with the maximum area that can be inscribed in a circle of radius r are Length = \(2\frac{r}{\sqrt{2}}\) and Width = \(2\frac{r}{\sqrt{2}}\). It is a square with side length \(\sqrt{2}r\), and a maximum area of \(2r^2\).

Step by step solution

01

Draw a diagram and label the dimensions

Draw a circle with radius r and inscribe a rectangle inside it. Label the length of the rectangle as 2x and the width as 2y, since the rectangle is inscribed in the circle the diagonal of the rectangle is the diameter of the circle.
02

Create the geometric relationship

Using the Pythagorean theorem, we can relate the dimensions of the rectangle to the radius of the circle: \(r^2 = (x^2 + y^2)\) Since length and width are positive, we know: \(x^2 = r^2 - y^2\) or \(x = \sqrt{r^2 - y^2}\)
03

Find the area function

The area A of the rectangle can be written as: \(A(x,y) = 2x \cdot 2y = 4xy\) Now, replace x with the expression we found in step 2 to write the area in terms of y only: \(A(y) = 4y \cdot \sqrt{r^2 - y^2}\)
04

Optimize the area function using calculus

To optimize the area function, find its critical points by taking the first derivative and setting it to zero. To find the first derivative, use the product rule and the chain rule: \(A'(y) = 4\sqrt{r^2-y^2} - \frac{4y^2}{\sqrt{r^2-y^2}}\) Set the derivative to zero to find the critical points: \(4\sqrt{r^2-y^2} = \frac{4y^2}{\sqrt{r^2-y^2}}\) Now square both sides to get rid of the square roots: \(16(r^2 - y^2) = 16y^4\) Now solve for y: \(y^4 - r^2y^2 + r^4 = 0\) This can be written as a quadratic equation in \(y^2\): \((y^2)^2 - r^2(y^2) + r^4 = 0\) Using quadratic formula to solve for \(y^2\): \(y^2 = \frac{r^2 \pm \sqrt{r^4 - 4r^4}}{2}\) \(y^2 = \frac{r^2}{2}\) So, \(y=\frac{r}{\sqrt{2}}\)
05

Determine the dimensions of the rectangle

Now we found the optimal value of y. Substitute it back into the equation for x in step 2 to find the optimal value of x: \(x = \sqrt{r^2 - (\frac{r}{\sqrt{2}})^2} = \frac{r}{\sqrt{2}}\) Therefore, the dimensions of the rectangle with the largest area that can be inscribed in a circle of radius r are: Length 2x = 2\(\frac{r}{\sqrt{2}}\) and Width 2y = 2\(\frac{r}{\sqrt{2}}\) This is a square with side length \(\sqrt{2}r\), and a maximum area of \(2r^2\).

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!

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

In an apple orchard there are 30 trees per acre, and the average yield per tree is 400 apples. For each additional tree planted per acre, the average yield per tree is reduced by 10 apples. How many trees per acre will maximize the crop?

Two vertical poles 15 and \(20 \mathrm{ft}\) above the ground are to be reinforced by a wire connected to the top of the poles and tied to a stake driven in the ground between the poles. Locate the stake with respect to one of the poles if the poles are \(21 \mathrm{ft}\). apart, and a minimum amount of wire is to be used.

In a given semi-circle of radius 2, in which a rectangle is to be inscribed, what are the dimensions for maximum area of the rectangle?

A piece of wire 20 inches long is to be cut and made into a rectangular frame. What dimensions should be chosen so that the area of the rectangle enclosed is maximal?

The velocity of a particle in feet per second is given by $$ \mathrm{v}(\mathrm{t})=\mathrm{t}^{2}-4 \mathrm{t}+5, \mathrm{t} \geqq 0 $$ where \(t\) is in seconds. Find the time at which the velocity is a relative maximum or a relative minimum.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

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.