a. Answer The y coordinate of point P is − x + 1 Explanation Calculation: Given the hypotenuse of an isosceles triangle is 2 units longer The numbers are x and y 0 < x < 20 x + y = 20 y = 20 − x The equation of straight line A B is y = − x + 1 Thus, the y coordinate of point P is − x + 1 b. To determine Express the area of the rectangle in learns of x . Answer Area of the rectangle is − 2 x 2 + 2 x Explanation Calculation: Given the hypotenuse of an isosceles triangle is 2 units longer The numbers are x and y 0 < x < 20 x + y = 20 y = 20 − x Area of the rectangle A ( x ) = 2 x ( − x + 1 ) = − 2 x 2 + 2 x Thus, Area of the rectangle is − 2 x 2 + 2 x c. To determine Find the largest area of the area rectangle in learns of x . Answer The maximum area of the rectangle is ( 1 2 ) Explanation Calculation: Given the hypotenuse of an isosceles triangle is 2 units longer The numbers are x and y 0 < x < 20 x + y = 20 y = 20 − x Area of the rectangle A = − 2 x 2 + 2 x d A d x = − 4 x + 2 For any function f ( x ) the condition of minimum and maximum is obtained by find f ′ ( x ) = 0 , Critical points and end points Critical points occur at A ′ = 0 − 4 + 2 = 0 x = 2 4 = 1 2 solve it graphically, it has critical point occur at x = 1 2 here d A d x > 0 for 0< x < 1 2 and d A d x < 0 for 1 2 < x < 1 this corresponds to maximum area. For x = 1 2 y = 2 x = 1 A = − 2 x 2 + 2 x = − ( 2 ) ( 1 2 ) 2 +2 ( 1 2 ) = ( 1 2 ) Graphical support: Thus, the maximum area of the rectangle is ( 1 2 )

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.4, Problem 5E

To determine

Find the y coordinate of a point P of a rectangle inscribed in an isosceles right triangle.

Expert Solution

Answer to Problem 5E

The y coordinate of point P is −x+1

Explanation of Solution

Calculation:

Given the hypotenuse of an isosceles triangle is 2 units longer

The numbers are x and y 0<x<20

x+y=20y=20−x

The equation of straight line AB is y=−x+1

Thus, the y coordinate of point P is −x+1

To determine

Express the area of the rectangle in learns of x .

Expert Solution

Answer to Problem 5E

Area of the rectangle is −2x2+2x

Explanation of Solution

Calculation:

Given the hypotenuse of an isosceles triangle is 2 units longer

The numbers are x and y 0<x<20

x+y=20y=20−x

Area of the rectangle

A(x)=2x(−x+1) =−2x2+2x

Thus, Area of the rectangle is −2x2+2x

To determine

Find the largest area of the area rectangle in learns of x .

Expert Solution

Answer to Problem 5E

The maximum area of the rectangle is (12)

Explanation of Solution

Calculation:

Given the hypotenuse of an isosceles triangle is 2 units longer

The numbers are x and y 0<x<20

x+y=20y=20−x

Area of the rectangle

A=−2x2+2xdAdx=−4x+2

For any function f(x) the condition of minimum and maximum is obtained by find f′(x)=0 ,

Critical points and end points

Critical points occur at

A′=0−4+2=0 x=24 =12

solve it graphically, it has critical point occur at x=12

here dAdx>0 for 0< x<12anddAdx<0 for12< x<1 this corresponds to maximum area.

For

x=12 y=2x=1 A =−2x2+2x =−(2)(12)2+2 (12) =(12)

Graphical support:

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.4, Problem 5E

Thus, the maximum area of the rectangle is (12)

Chapter 5.4, Problem 4E

Chapter 5.4, Problem 6E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

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