Ch. 4 - Applications of the Derivative

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 4 - Applications of the Derivative

Problem 4.5.11

Chapter 4, Problem 4.5.11

Maximum-area rectangles Of all rectangles with a perimeter of 10, which one has the maximum area? (Give the dimensions.)

Verified step by step guidance

Start by understanding the relationship between the perimeter and the dimensions of a rectangle. The perimeter P of a rectangle with length l and width w is given by the formula:

P = 2 (l + w)

. Since the perimeter is 10, we have:

2 (l + w) = 10

Solve the perimeter equation for one of the variables, say w. From

2 (l + w) = 10

, we get:

l + w = 5

. Therefore,

w = 5 - l

Express the area A of the rectangle in terms of l. The area A is given by:

A = l \times w

. Substitute the expression for w from the previous step:

A = l \times (5 - l)

Simplify the expression for the area:

A = 5 l - l^{2}

. This is a quadratic function in terms of l, which can be written as:

A = - l^{2} + 5 l

Find the value of l that maximizes the area. Since the quadratic function

A = - l^{2} + 5 l

is a downward-opening parabola, its maximum value occurs at the vertex. The vertex of a quadratic function

ax + bx + c

is given by

- b / 2 a

. Here,

a = - 1

and

b = 5

, so calculate

- 5 / (2 \times - 1)

to find the optimal length l.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Perimeter

Perimeter is the total distance around a two-dimensional shape. For rectangles, the perimeter (P) is calculated using the formula P = 2(length + width). In this problem, the perimeter is fixed at 10, which means the sum of the length and width must equal 5, allowing us to express one dimension in terms of the other.

Area of a Rectangle

The area of a rectangle is calculated by multiplying its length (l) by its width (w), expressed as A = l * w. To find the rectangle with the maximum area under a fixed perimeter, we need to express the area in terms of a single variable, which can be done by substituting the width with a function of the length derived from the perimeter constraint.

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function. In this context, we will use techniques such as taking the derivative of the area function and setting it to zero to find critical points. This will help us determine the dimensions of the rectangle that yield the maximum area while adhering to the given perimeter constraint.

Recommended video:

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Intro to Applied Optimization: Maximizing Area

Related Practice

Textbook Question

Differentials Consider the following functions and express the relationship between a small change in x and the corresponding change in y in the form dy = f'(x)dx.

f(x) = (x+4)/(4-x)

Textbook Question

Finding antiderivatives. Find all the antiderivatives of the following functions. Check your work by taking derivatives.

g(s) = 1 / (s² + 1)

Textbook Question

Suppose f is differentiable on (-∞,∞), f(1) = 2, and f'(1) = 3. Find the linear approximation to f at x = 1 and use it to approximate f (1.1).

Textbook Question

Graphing functions Use the guidelines of this section to make a complete graph of f.

f(x) = e⁻ˣ²/₂

Textbook Question

Absolute maxima and minima Determine the location and value of the absolute extreme values of ƒ on the given interval, if they exist.

ƒ(x) = x/(x²+9)⁵ on [-2,2]

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Textbook Question

{Use of Tech} Finding roots with Newton’s method For the given function f and initial approximation x₀, use Newton’s method to approximate a root of f. Stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. Show your work by making a table similar to that in Example 1.

f(x) = x² - 10; x₀ = 3

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