BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071
BuyFind

Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

Solutions

Chapter 2, Problem 28P
To determine

To find: The dimension of a rectangle for which its area is maximum and two of its vertices are on x-axis and two vertices are on y=8x2

Expert Solution

Answer to Problem 28P

The length of inscribed rectangle is 3.26 and height is 5.35 .

Explanation of Solution

Given:

The given figure is,

Precalculus: Mathematics for Calculus - 6th Edition, Chapter 2, Problem 28P , additional homework tip  1

Figure (1)

Calculation:

Area of rectangle is,

A=(length)(width)=(l)(w)

From Figure (1) length is 2x and width is h .

Substitute 2x for l and y for w in equation (1)

A=(2x)(y)=2xy

Vertices of the rectangle are on y=8x2 .

Summarize the information in a table as shown below.

In Words In Algebra
Area lw
Length of side rectangle. 2x
Height of the rectangle 8x2

Use the information in the table and model the area in terms of x .

A=(l)(w)A(x)=(2x)(8x2)[y=8x2]=16x2x3=2(8xx3)

The function that models the area of given rectangle is,

A(x)=2(8xx3)

Sketch the graph of A(x) in first quadrant (because area and dimensions of rectangle is non-negative) using graphing calculator as shown below.

Precalculus: Mathematics for Calculus - 6th Edition, Chapter 2, Problem 28P , additional homework tip  2

Figure (2)

Observe from Figure (2) that function attains a maximum value when x is 1.63.

Substitute 1.63 for x in y=8x2 and solve for y.

y=8(1.63)2=82.65=5.35

The length of rectangle for maximum area is,

l=2x

Substitute 1.63 for x in above equation

l=3.26

Thus, the length of inscribed rectangle is 3.26 and height is 5.35 .

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