a. Answer The dimension of width w = 6 and diameter is d = 6 3 in. Explanation Given Maximum of S is at w = 4 3 . Calculation: Pythagoras’ theorem, w 2 + d 2 = 144 S ( w ) = w d 2 = w ( 144 − w 2 ) 1.5 use product rule and chain rule S ′ ( w ) = d w d w ( 144 − w 2 ) 1.5 + w d ( 144 − w 2 ) 1.5 d w 1 ⋅ ( 144 − w 2 ) 1.5 + w ⋅ 1.5 ( 144 − w 2 ) ( 1.5 − 1 ) ⋅ d ( 144 − w 2 ) d w = ( 144 − w 2 ) 1.5 + 1.5 w ( 144 − w 2 ) ( 0.5 ) ( − 2 w ) = ( 144 − w 2 ) ( 0.5 ) ( 144 − w 2 − 3 w 2 ) = 144 − w 2 ( 144 − 4 w 2 ) S ′ ( w ) = 0 local extrema 144 − w 2 ( 144 − 4 w 2 ) = 0 w = ± 12 , ± 6 d ( 6 ) = 6 3 , S ( 6 ) = 6 4 ∗ 3 3 d ( 6 ) = 0 , S ( 12 ) = 0 S ( 0 ) = 0 < S ( 6 ) Thus, the width is w = 6 , and d = 6 3 . b. To determine Check your answer in part (a) Answer s = w ( 144 − w 2 ) 3 2 Explanation Given Maximum of S is at w = 4 3 . Calculation: Maximum ofS is at w = 4 3 , d = 6 3 from part (a) k = 1 S = k w d 3 = w d 3 12 2 = w 2 + d 2 d = 144 − w 2 s = w ( 144 − w 2 ) 3 2 c. To determine Compare the graphs. Answer Graph of S ( w ) in light blue. Graph of S ( d ) in purple. Explanation Given Maximum of S is at w = 4 3 . Calculation: Rewrite the strength in terms of d , k = 1 S = k w d 3 = w d 3 S ( w ) = w ( 144 − w 2 ) 3 / 2 12 2 = w 2 + d 2 w = 144 − d 2 S ( d ) = d 3 ( 144 − d 2 ) Graph of S ( w ) in light blue. Graph of S ( d ) in purple. Here the same graohs with k = 12 Graph of S ( w ) in light blue. The maximum occurs at w = 6 which agrees with the answer from part (a).

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

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Differential Equations

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Question

Chapter 5.4, Problem 38E

To determine

To find the dimensions of the stiffest beam.

Expert Solution

Answer to Problem 38E

The dimension of width w=6 and diameter is d=63 in.

Explanation of Solution

Given Maximum of S is at w=43.

Calculation:

Pythagoras’ theorem,

w2+d2=144S(w)=wd2 =w(144−w2)1.5use product rule and chain ruleS′(w)=dwdw(144−w2)1.5+wd(144−w2)1.5dw1⋅(144−w2)1.5+w⋅1.5(144−w2)(1.5−1)⋅d(144−w2)dw=(144−w2)1.5+1.5w(144−w2)(0.5)(−2w)=(144−w2)(0.5)(144−w2−3w2)=144−w2(144−4w2)S′(w)=0 local extrema144−w2(144−4w2)=0w=±12,±6d(6)=63,S(6)=64∗33d(6)=0,S(12)=0S(0)=0<S(6)

Thus, the width is w=6 , and d=63 .

To determine

Check your answer in part (a)

Expert Solution

Answer to Problem 38E

s=w(144−w2)32

Explanation of Solution

Given Maximum of S is at w=43.

Calculation:

Maximum ofS is at w=43 , d=63 from part (a)

k=1S=kwd3 =wd3122=w2+d2d=144−w2s=w(144−w2)32

To determine

Compare the graphs.

Expert Solution

Answer to Problem 38E

Graph of S(w) in light blue. Graph of S(d) in purple.

Explanation of Solution

Given Maximum of S is at w=43.

Calculation:

Rewrite the strength in terms of d ,

k=1S=kwd3 =wd3S(w)=w(144−w2)3/2122=w2+d2w=144−d2S(d)=d3(144−d2)

Graph of S(w) in light blue.

Graph of S(d) in purple.

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.4, Problem 38E , additional homework tip 1

Here the same graohs with k=12

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.4, Problem 38E , additional homework tip 2

Graph of S(w) in light blue.

The maximum occurs at w=6 which agrees with the answer from part (a).

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy), Chapter 5.4, Problem 38E , additional homework tip 3

Chapter 5.4, Problem 37E

Chapter 5.4, Problem 39E

Chapter 5 Solutions

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

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

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