Explanation Given information: The function is f ( x , y ) = y 2 − x y − 3 y + 2 x . The square region is enclosed by 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 ) = y 2 − x y − 3 y + 2 x (1) Draw the square region bounded by 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 AB : Refer to Figure 1. For segment AB , x = − 2 . Substitute − 2 for x in Equation (1). f ( − 2 , y ) = y 2 + 2 y − 3 y − 4 = y 2 − y − 4 Thus, f ( − 2 , y ) = y 2 − y − 4 for the interval − 2 ≤ y ≤ 2 . Differentiate with respect to y and equate to 0. f ' ( − 2 , y ) = 2 y − 1 = 0 y = 1 2 The point ( − 2 , 1 2 ) lies in the region. Substitute 2 for x and 1 2 for y in Equation (1). f ( − 2 , 1 2 ) = 1 4 + 1 − 3 2 − 4 = 1 + 4 − 6 − 16 4 = − 17 4 Thus, ( − 2 , 1 2 ) is an interior critical point with the value of − 17 4 . The endpoints of segment AB are ( − 2 , − 2 ) and ( 2 , 2 ) . Find the value at the endpoints as follows: Substitute − 2 for x and − 2 for y in Equation (1). f ( − 2 , − 2 ) = 4 − 4 + 6 − 4 = 2 Substitute 2 for x and 2 for y in Equation (1). f ( 2 , 2 ) = 4 − 4 − 6 + 4 = − 2 Consider Segment BC : Refer to Figure 1. For segment BC , y = 2 . Substitute 2 for y in Equation (1). f ( x , 2 ) = 4 − 2 x − 6 + 2 x = − 2 Thus, f ( x , 2 ) = 2 for the interval − 2 ≤ x ≤ 2 . Differentiate once with respect to x and equate to zero. f ' ( x , 0 ) = 0 There is no critical point in the interior region of BC . The endpoints of segment BC are ( − 2 , 2 ) and ( 2 , 2 ) . Find the value at the endpoints as follows: Substitute − 2 for x and 2 for y in Equation (1). f ( − 2 , 2 ) = 4 + 4 − 6 − 4 = − 2 Substitute 2 for x and 2 for y in Equation (1). f ( 2 , 2 ) = 4 − 4 − 6 + 4 = − 2 Consider Segment CD : Refer to Figure 1

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 73PE

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Calculate the absolute maxima and minima of the given function.

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Chapter 14, Problem 72PE

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