Explanation Given: The function is f ( x , y ) = x 2 − y 2 − 2 x + 4 y + 1 (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 : The first derivatives of the given function is computed as follows, Compute the partial derivative of the equation (1) with respect to x . ∂ ∂ x f ( x , y ) = ∂ ∂ x ( x 2 − y 2 − 2 x + 4 y + 1 ) = 2 x − 2 Thus, ∂ ∂ x f ( x , y ) = 2 x − 2 (2) Compute the partial derivative of the equation (1) with respect to y . ∂ ∂ y f ( x , y ) = ∂ ∂ y ( x 2 − y 2 − 2 x + 4 y + 1 ) = − 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 . That is, 2 x − 2 = 0 and − 2 y + 4 = 0 . Solve the equation 2 x − 2 = 0 and obtain the value of x . 2 x − 2 = 0 x = 2 2 x = 1 Similarly, the value of y is obtained as follows, − 2 y + 4 = 0 y = − 4 − 2 y = 2 Thus, the critical point of the given function is ( 1 , 2 )

Applied Calculus for the Managerial, Life, and Social Sciences

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ISBN: 9781305887831

Author: Tan

Publisher: CENGAGE CO

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Chapter 8.3, Problem 3E

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

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Chapter 8.3, Problem 2E

Chapter 8.3, Problem 4E

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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. Compute g(1, 1, 1), g(1,...Ch. 8.1 - Prob. 10E Ch. 8.1 - Prob. 11E Ch. 8.1 - Prob. 12E Ch. 8.1 - In Exercises 11-18, find the domain of the...Ch. 8.1 - Prob. 14E Ch. 8.1 - In Exercises 11-18, find the domain of the...Ch. 8.1 - Prob. 16E Ch. 8.1 - In Exercises 11-18, find the domain of the...Ch. 8.1 - Prob. 18E Ch. 8.1 - Prob. 19E Ch. 8.1 - Prob. 20E Ch. 8.1 - In Exercises 19-24, sketch the level curves of the...Ch. 8.1 - Prob. 22E Ch. 8.1 - In Exercises 1924, sketch the level curves of the...Ch. 8.1 - Prob. 24E Ch. 8.1 - Find an equation of the level curve of f (x, y) =...Ch. 8.1 - Prob. 26E Ch. 8.1 - Prob. 27E Ch. 8.1 - Prob. 28E Ch. 8.1 - Prob. 29E Ch. 8.1 - Prob. 30E Ch. 8.1 - Prob. 31E Ch. 8.1 - Prob. 32E Ch. 8.1 - Prob. 33E Ch. 8.1 - Prob. 34E Ch. 8.1 - Prob. 35E Ch. 8.1 - POISEITLLES LAW Poiseuilles Law states that the...Ch. 8.1 - COST FUNCTION FOR A LOUDSPEAKER SYSTEM Acrosonic...Ch. 8.1 - Prob. 38E Ch. 8.1 - REVENUE FUNCTIONS FOR DESKS Country Workshop...Ch. 8.1 - Prob. 40E Ch. 8.1 - 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The critical point and relative extrema of the function; Classify the nature of each point.

Solution Summary: The author explains the critical point and relative extrema of the function. The saddle point at (1,2) is 4.

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