(a) Explanation Given information: Consider the function f ( x , y ) = x 2 + y 2 + 2 x y − x − y + 1 over the square 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 . Procedure: Use the following procedure to find out the 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 called as absolute maximum and minimum value of the function. Here, the function is indicated as D . Calculation: Provide the function as shown below: f ( x , y ) = x 2 + y 2 + 2 x y − x − y + 1 (1) Draw the bounded region in the first quadrant using the lines x = 0 , y = 1 , and y = 0 as shown in Figure 1. Find the interior points: Consider the function D is differentiable. Hence, f x = f y = 0 . Find the first order partial derivative with respect to x , f x . f x = 2 x + 2 y − 1 = 0 2 x + 2 y − 1 = 0 Find the first order partial derivative with respect to y , f y . f y = 2 y + 2 x − 1 = 0 2 x + 2 y = 1 x + y = 1 2 Find the boundary points: Consider the segment OC from the figure 1. From the segment OC , x = 0 . Show the function expressed as follows: g ( y ) = f ( 0 , y ) . Find f ( 0 , y ) using Equation 1. f ( 0 , y ) = 0 2 + y 2 + 2 ( 0 ) y − 0 − y + 1 = y 2 − y + 1 For the interval 0 ≤ y ≤ 1 , g ′ ( y ) = 0 . g ′ ( y ) = 0 2 y − 1 = 0 y = 1 2 The coordinate of the segment f ( 0 , 0 ) , f ( 0 , 1 2 ) , and f ( 0 , 1 ) . Find f ( 0 , 0 ) using Equation 1. f ( 0 , 0 ) = 0 2 + 0 2 + 2 ( 0 ) − 0 − 0 + 1 = 0 + 1 = 1 Find f ( 0 , 1 2 ) using Equation 1. f ( 0 , 1 2 ) = 0 2 + ( 1 2 ) 2 + 2 ( 0 ) ( 1 2 ) − 0 − ( 1 2 ) + 1 = 0 + 1 4 − 1 2 + 1 = 1 − 2 + 4 4 = 3 4 Find f ( 0 , 1 ) using Equation 1. f ( 0 , 1 ) = 0 2 + 1 2 + 2 ( 0 ) ( 1 ) − 0 − 1 + 1 = 1 Segment CB : Consider the segment CB from the figure 1. From the segment CB , y = 1 . Show the function expressed as follows: g ( x ) = f ( x , 1 ) . Find f ( x , 1 ) using Equation 1. f ( x , 1 ) = x 2 + 1 + 2 x ( 1 ) − x − 1 + 1 = x 2 + x + 1 For the interval 0 ≤ x ≤ 1 , g ′ ( x ) = 0 (b) To determine Find the absolute maximum value of f .

Thomas' Calculus: Early Transcendentals in SI Units

14th Edition

ISBN: 9781292253114

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

Publisher: PEARSON

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Chapter 14.7, Problem 60E

(a)

To determine

Prove that the minimum value of f estimated along the line equation 2x+2y=1 in the square region and the absolute minimum value.

(b)

To determine

Find the absolute maximum value of f.

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Chapter 14.7, Problem 59E

Chapter 14.7, Problem 61E

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Prove that the minimum value of f estimated along the line equation 2 x + 2 y = 1 in the square region and the absolute minimum value.