(a) Explanation Given information: The function f ( x , y ) = x 2 + y 2 . The line segment x = t , y = 2 − 2 t . Calculation: Provide the function as follows: f ( x , y ) = x 2 + y 2 (1) Substitute t for x and 2 − 2 t for y in Equation (1). f ( t ) = t 2 + ( 2 − 2 t ) 2 = t 2 + 4 + 4 t 2 − 8 t = 5 t 2 − 8 t + 4 (2) Thus, the function reduced in to the function of one variable only, whose extreme points occurs at the end points of the domain or at the critical points. Calculate the critical points as shown below. Differentiate Equation (2) with respect to t . f ′ ( t ) = ∂ ∂ t ( 5 t 2 − 8 t + 4 ) = 10 t − 8 Equate first order partial derivative with respect to t , ∂ f ∂ t = 0 . 0 = 10 t − 8 t = 4 5 There is only one critical points. For the line, where a maximum or minimum value can occur at the critical point t = 4 5 . Calculate the absolute maximum and minimum value as shown below. Substitute 4 5 for t in Equation (2). f ( 4 5 ) = 5 ( 4 5 ) 2 − 8 ( 4 5 ) + 4 = 80 25 − 32 5 + 4 = 80 − 160 + 100 25 = 4 5 Thus, by second derivative test, the minimum value of the function f is 4 5 at t = 4 5 and the function has no maximum value on the line (b) To determine Calculate the absolute maximum and minimum value for the function.

14th Edition

ISBN: 9780135430903

Author: Hass

Publisher: PEARSON

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Chapter 14.7, Problem 66E

(a)

To determine

Calculate the absolute maximum and minimum value for the function.

(b)

To determine

Calculate the absolute maximum and minimum value for the function.

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Chapter 14.7, Problem 65E

Chapter 14.7, Problem 67E

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