Explanation Given: The function is f ( x , y ) = 2 x 3 + y 2 − 9 x 2 − 4 y + 12 x − 2 (1) Result used: Second derivative test “Let D ( x , y ) = f x x ( x , y ) f y y ( x , y ) − f 2 x y ( x , y ) . Then, check the values of D ( a , b ) as below. (a) If D ( a , b ) > 0 and f x x < 0 then the function f has a relative maximum at the critical point ( a , b ) . (b) If D ( a , b ) > 0 and f x x > 0 then the function f has a relative minimum at the critical point ( a , b ) . (c) If D ( a , b ) < 0 then the function f has a no relative extrema at the critical point ( a , b ) . The point ( a , b , f ( a , b ) ) is said to be a saddle point. (d) If D ( a , b ) = 0 then there will be no result. It is to be noted that in (a) and (b) f x x and f y y must have the same sign to satisfy the condition D ( a , b ) > 0 ”. Calculation: Obtain the critical point of the function : Compute the partial derivative of the equation (1) with respect to x . ∂ ∂ x f ( x , y ) = ∂ ∂ x ( 2 x 3 + y 2 − 9 x 2 − 4 y + 12 x − 2 ) = ∂ ∂ x ( 2 x 3 ) + ∂ ∂ x ( y 2 ) − ∂ ∂ x ( 9 x 2 ) − ∂ ∂ x ( 4 y ) + ∂ ∂ x ( 12 x ) − ∂ ∂ x ( 2 ) = 6 x 2 + 0 − 18 x − 0 + 12 ( 1 ) − 0 = 6 x 2 − 18 x + 12 Thus, ∂ ∂ x f ( x , y ) = 6 x 2 − 18 x + 12 (2) Compute the partial derivative of the equation (1) with respect to y . ∂ ∂ y f ( x , y ) = ∂ ∂ y ( 2 x 3 + y 2 − 9 x 2 − 4 y + 12 x − 2 ) = ∂ ∂ y ( 2 x 3 ) + ∂ ∂ y ( y 2 ) − ∂ ∂ y ( 9 x 2 ) − ∂ ∂ y ( 4 y ) + ∂ ∂ y ( 12 x ) − ∂ ∂ y ( 2 ) = 0 + 2 y − 0 − 4 ( 1 ) + 0 − 0 = 2 y − 4 Hence, ∂ ∂ y f ( x , y ) = 2 y − 4 (3) Obtain the critical points by equating f x ( x , y ) = 0 and f y ( x , y ) = 0 . Then the equation (3) becomes, 6 x 2 − 18 x + 12 = 0 6 x 2 − 18 x = − 12 (4) And equation (4) becomes, 2 y − 4 = 0 2 y = 4 (5) Obtain the value of y from equation (5). 2 y = 4 y = 4 2 y = 2 Thus, the value of y is 2. Obtain the value of x from equation (4). 6 x 2 − 18 x + 12 ( x − 2 ) ( x − 1 ) = 0 Thus, the values of x are x = 2 and x = 1

Applied Calculus for the Managerial, Life, and Social Sciences

10th Edition

ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 8.3, Problem 7E

To determine

To find: The critical points and relative extrema of the function; Classify the nature of the function of each point, if possible.

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Chapter 8.3, Problem 6E

Chapter 8.3, Problem 8E

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Calculus III May I please have some elaborations on Example 2 part a? Thank you.

1. A bicyclist is riding their bike along the Chicago Lakefront Trail. The velocity (in feet per second) of the bicyclist is recorded below. Use (a) Simpson's Rule, and (b) the Trapezoidal Rule to estimate the total distance the bicyclist traveled during the 8-second period. t 0 2 4 6 8 V 10 15 12 10 16 2. Find the midpoint rule approximation for (a) n = 4 +5 x²dx using n subintervals. 1° 2 (b) n = 8 36 32 28 36 32 28 24 24 20 20 16 16 12 8- 4 1 2 3 4 5 6 12 8 4 1 2 3 4 5 6

= 5 37 A 4 8 0.5 06 9

Chapter 8 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences

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Ch. 8.1 - Prob. 8E Ch. 8.1 - Let g(r, s, t) = res/t. 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Prob. 30E Ch. 8.5 - Prob. 31E Ch. 8.5 - Prob. 33E Ch. 8.5 - Prob. 34E Ch. 8.5 - Prob. 36E Ch. 8.5 - Prob. 38E Ch. 8.5 - Prob. 39E Ch. 8.5 - Prob. 40E Ch. 8.5 - Prob. 41E Ch. 8.5 - Prob. 42E Ch. 8.6 - Prob. 1CQ Ch. 8.6 - Prob. 2CQ Ch. 8.6 - Prob. 1E Ch. 8.6 - Prob. 2E Ch. 8.6 - Prob. 3E Ch. 8.6 - Prob. 4E Ch. 8.6 - Prob. 5E Ch. 8.6 - Prob. 6E Ch. 8.6 - Prob. 7E Ch. 8.6 - Prob. 8E Ch. 8.6 - Prob. 9E Ch. 8.6 - Prob. 10E Ch. 8.6 - Prob. 11E Ch. 8.6 - Prob. 12E Ch. 8.6 - Prob. 13E Ch. 8.6 - Prob. 14E Ch. 8.6 - In Exercises 318, find the total differential of...Ch. 8.6 - Prob. 16E Ch. 8.6 - Prob. 17E Ch. 8.6 - Prob. 18E Ch. 8.6 - Prob. 19E Ch. 8.6 - Prob. 20E Ch. 8.6 - Prob. 21E Ch. 8.6 - Prob. 22E Ch. 8.6 - Prob. 23E Ch. 8.6 - Prob. 24E Ch. 8.6 - Prob. 25E Ch. 8.6 - Prob. 26E Ch. 8.6 - Prob. 27E Ch. 8.6 - Prob. 28E Ch. 8.6 - Prob. 29E Ch. 8.6 - Prob. 30E Ch. 8.6 - Prob. 31E Ch. 8.6 - Prob. 32E Ch. 8.6 - Prob. 33E Ch. 8.6 - Prob. 34E Ch. 8.6 - Prob. 35E Ch. 8.6 - Prob. 36E Ch. 8.6 - Prob. 37E Ch. 8.6 - 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The critical points and relative extrema of the function; Classify the nature of the function of each point, if possible.

Solution Summary: The author explains the critical points and relative extrema of the function.

To find: The critical points and relative extrema of the function; Classify the nature of the function of each point, if possible.

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Chapter 8 Solutions