Chapter 8.4, Problem 18E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# A rectangle has an area of 36 in2. What is the limit (smallest possible value) of the perimeter of the rectangle?

To determine

To Find:

The limit of the perimeter of a rectangle, has an area of 36 in2.

Explanation

Given:

The area of a rectangle is 36 in2.

Formula:

Let x represents one side and y represents the other side of the rectangle.

The area of the rectangle A=xy.

The perimeter of a rectangle is P=2(x+y).

Calculation:

The area of the rectangle is,

xy=36.

y=36x.

Substitute the y value in perimeter of a rectangle,

Then, P=2(x+36x).

P(x)=2x+72x.

To find the smallest perimeter, P'(x)=0.

P'(x)=2âˆ’72x2.

The smallest perimeter is P'(x)=2âˆ’72x2=0

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