   Chapter 8.2, Problem 37E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# Examine several rectangles, each with a perimeter of 40 in., and find the dimensions of the rectangle that has the largest area. What type of figure has the largest area?

To determine

To Find:

The dimensions of the rectangle that has the largest area provided its perimeter.

Explanation

Formula Used:

The perimeter of a polygon is the sum of the lengths of all sides of the polygon.

Calculation:

It is given that the perimeter of the rectangle 40 in.

Let 'x' be the length and 'y' be the height of the rectangle. Then the perimeter becomes

2x+y=40

Dividing by '2',

x+y=20(1)

Let’s find the rectangle that has largest area, given this perimeter.

Applying the area formula, A=x×y(2)

The height of the rectangle can be written as y=20-x…(3)

Substitute the value of height from equation (3) in equation (2)

A=x(20-x)

A=20x-x2

A=-x2+20x

Therefore the provided area is quadratic. Let’s find the maximum area. Since the above area equation is a negative quadratic, then it graphs as an upside-down parabola, so the vertex is the maximum.

Let’s find the vertex point.

The equation of the quadratic, in y= ax2+ bx+c.

The vertex of a parabola is the point (h, k), where h = b2a

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