Answer The width of the football field is 160 feet. The maximum height of the surface of the field is 1.4976 feet. Explanation Given Information: The cross section of a field is modelled by y = − 0.000234 x ( x − 160 ) where x and y are measured in feet. Calculation: To find the width of the football field first find the x-intercepts and then find the difference between them to find the distance. Substitute y = 0 . y = − 0.000234 x ( x − 160 ) 0 = − 0.000234 x ( x − 160 ) x = 0 , 160 The distance between the lowest and the highest x-intercept is, D = 160 − 0 = 160 Here, D is the width of the football field. The maximum height of the surface of the field is given by the y-coordinate of the vertex. The x -coordinate of the vertex of a parabola of the equation y = a x 2 + b x + c is , x = − b 2 a The equation y = − 0.000234 x ( x − 160 ) in the form y = a x 2 + b x + c is, y = − 0.000234 x ( x − 160 ) = − 0.000234 x 2 + 0.03744 On comparing, a = − 0.000234 b = 0.03744 c = 0 The x -coordinate is, x = − 0.03744 2 ( − 0.000234 ) = 37440 2 × 234 = 37440 468 = 80 The maximum height therefore is, y = − 0.000234 x ( x − 160 ) = − 0.000234 × 80 × ( 80 − 160 ) = 0.000234 × 6400 = 1.4976

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

15th Edition

ISBN: 9781608408405

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Cengage Learning

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Maximization

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Question

Chapter 2.2, Problem 66E

To determine

To find: The width of the football field and the maximum height of the surface of the field.

Expert Solution & Answer

Answer to Problem 66E

The width of the football field is 160 feet.

The maximum height of the surface of the field is 1.4976 feet.

Explanation of Solution

Given Information: The cross section of a field is modelled by y=−0.000234x(x−160) where x and y are measured in feet.

Calculation:

To find the width of the football field first find the x-intercepts and then find the difference between them to find the distance. Substitute y=0 .

y=−0.000234x(x−160)0=−0.000234x(x−160)x=0,160

The distance between the lowest and the highest x-intercept is,

D=160−0=160

Here, D is the width of the football field.

The maximum height of the surface of the field is given by the y-coordinate of the vertex.

The x -coordinate of the vertex of a parabola of the equation y=ax2+bx+c is ,

x=−b2a

The equation y=−0.000234x(x−160) in the form y=ax2+bx+c is,

y=−0.000234x(x−160)=−0.000234x2+0.03744

On comparing,

a=−0.000234b=0.03744c=0

The x -coordinate is,

x=−0.037442(−0.000234)=374402×234=37440468=80

The maximum height therefore is,

y=−0.000234x(x−160)=−0.000234×80×(80−160)=0.000234×6400=1.4976

Chapter 2.2, Problem 65E

Chapter 2.2, Problem 67E

Chapter 2 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

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