   Chapter 4.7, Problem 26E

Chapter
Section
Textbook Problem

Find the area of the largest rectangle that can be inscribed in the ellipse x2/a2 + y2/b2 = 1.

To determine

To find: The area of the largest rectangle that can be inscribed in the ellipse x2a2+y2b2=1 .

Explanation

Given:

The ellipse is x2a2+y2b2=1 .

Calculation:

The ellipse is 2a wide and 2b tall.

In Figure 1, x is half the width of the rectangle and y is half the length.

Hence the length of the rectangle is 2y and the width of the rectangle is 2x .

Therefore, the area is, A=(2x)(2y)=4xy

From the ellipse,

x2a2+y2b2=1y2b2=1x2a2yb=(1x2a2)y=ba(a2x2)

Substitute y=ba(a2x2) in A=4xy ,

A=4x(ba(a2x2))=4bxa(a2x2)

Differentiate A with respect to x,

For critical points, dAdx=0 .

4ba(a22x2(a2x2))=0a22x2(a2x2)=0a22x2=0x2=a

Simplify further as follows,

x=a2

Differentiate dAdx with respect x,

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