Explanation Given: Length of AB =3. Calculation: Let B C = 2 and A D = 5 . ∠ C P D = θ . In Figure 1, Let, ∠ B P C = α and ∠ A P D = β . Let B P = x . Hence, A P = 3 − x . In triangle Δ P B C , tan α = 2 x α = tan − 1 ( 2 x ) In triangle Δ P A D , tan β = 5 3 − x β = tan − 1 ( 5 3 − x ) Now, θ + α + β = 180 θ = 180 − α − β Substitute α = tan − 1 ( 2 x ) and β = tan − 1 ( 5 3 − x ) in θ , θ = 180 − tan − 1 ( 2 x ) − tan − 1 ( 5 3 − x ) Differentiate θ with respect to x , θ ′ = − ( 1 1 + ( 2 x ) 2 × ( − 2 x 2 ) ) − ( 1 1 + ( 5 3 − x ) 2 × 5 ( 3 − x ) 2 ) = 2 4 + x 2 − 5 x 2 − 6 x + 34 For critical points, θ ′ = 0 2 4 + x 2 − 5

Bundle: Single Variable Calculus: Early Transcendentals, Loose-leaf Version, 8th + Webassign Printed Access Card For Calculus, Multi-term Courses, Life Of Edition

18th Edition

ISBN: 9780357008034

Author: Stewart

Publisher: CENGAGE L

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Chapter 4.7, Problem 77E

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To find: The position of P on the line segment AB so that $θ$ is maximum.

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Chapter 4.7, Problem 76E

Chapter 4.7, Problem 78E

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Bundle: Single Variable Calculus: Early Transcendentals, Loose-leaf Version, 8th + Webassign Printed Access Card For Calculus, Multi-term Courses, Life Of Edition

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