Explanation Given information: The function is f ( x , y ) = 2 x + 2 y − x 2 − y 2 . The square region in the first quadrant is bounded by the coordinate axes and the lines x = 2 and y = 2 . Use the following procedure to find out the absolute maxima and minima: Find the interior points of the region R where the function f may have local maxima and minima and evaluate f at these points. It is called critical point of f . Find the maximum and minimum value of f from the list of points that is said to be absolute maximum and minimum value of the function. Calculation: Show the function f as follows: f ( x , y ) = 2 x + 2 y − x 2 − y 2 (1) Draw the square region bounded in the first quadrant by the coordinate axes and the lines x = 2 and y = 2 as shown in Figure 1. Let the function f is differentiable. Hence, f x = f y = 0 . Consider Segment OA : Refer to Figure 1. For segment OA , x = 0 . Substitute 0 for x in Equation (1). f ( 0 , y ) = 0 + 2 y − 0 − y 2 = 2 y − y 2 Thus, f ( 0 , y ) = 2 y − y 2 for the interval 0 ≤ y ≤ 2 . Differentiate with respect to y and equate to 0. f ' ( 0 , y ) = 2 − 2 y = 0 2 y = 2 y = 1 The point ( 0 , 1 ) lies in the region. Substitute 0 for x and 1 for y in Equation (1). f ( 0 , 1 ) = 0 + 2 − 0 − 1 = 1 Thus, ( 0 , 1 ) is an interior critical point with the value of 1 . The endpoints of segment OA are ( 0 , 0 ) and ( 0 , 2 ) . Find the value at the endpoints as follows: Substitute 0 for x and 0 for y in Equation (1). f ( 0 , 0 ) = 0 Substitute 0 for x and 2 for y in Equation (1). f ( 0 , 2 ) = 0 + 4 − 0 − 4 = 0 Consider Segment AB : Refer to Figure 1. For segment AB , y = 2 . Substitute 2 for y in Equation (1). f ( x , 2 ) = 2 x + 4 − x 2 − 4 = 2 x − x 2 Thus, f ( x , 2 ) = 2 x − x 2 for the interval 0 ≤ x ≤ 2 . Differentiate once with respect to x and equate to zero. f ' ( x , 0 ) = 2 − 2 x = 0 2 x = 2 x = 1 The point ( 1 , 2 ) lies in the region. Substitute 1 for x and 2 for y in Equation (1). f ( 1 , 2 ) = 2 + 4 − 1 − 4 = 1 Thus, ( 1 , 2 ) is an interior critical point with the value of 1 . The endpoints of segment AB are ( 0 , 2 ) and ( 2 , 2 ) . Find the value at the endpoints as follows: Substitute 0 for x and 2 for y in Equation (1). f ( 0 , 2 ) = 0 + 4 − 0 − 4 = 0 Substitute 2 for x and 2 for y in Equation (1). f ( 2 , 2 ) = 4 + 4 − 4 − 4 = 0 Consider Segment BC : Refer to Figure 1. For segment BC , x = 2 . Substitute 2for x in Equation (1). f ( 2 , y ) = 4 + 2 y − 4 − y 2 = 2 y − y 2 Thus, f ( 2 , y ) = 2 y − y 2 for the interval 0 ≤ y ≤ 2

MyLab Math plus Pearson eText -- Standalone Access Card -- for Thomas' Calculus: Early Transcendentals (14th Edition)

14th Edition

ISBN: 9780134764528

Author: Joel R. Hass, Christopher E. Heil, Maurice D. Weir

Publisher: PEARSON

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Chapter 14, Problem 74PE

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Calculate the absolute maxima and minima of the given function.

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Chapter 14, Problem 73PE

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MyLab Math plus Pearson eText -- Standalone Access Card -- for Thomas' Calculus: Early Transcendentals (14th Edition)

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