Explanation Given information: The function is, f ( x , y ) = x 2 + 3 y 2 − 4 x + 2 y − 3 Calculation: Consider the function, f ( x , y ) = x 2 + 3 y 2 − 4 x + 2 y − 3 ............ (1) Partial derivative of equation (1) with respect to x . f x = d d x ( x 2 + 3 y 2 − 4 x + 2 y − 3 ) 0 = 2 x − 4 x = 2 Partial derivative of equation (1) with respect to y . f y = d d y ( x 2 + 3 y 2 − 4 x + 2 y − 3 ) 0 = 6 y + 2 y = − 1 3 So, the critical point is ( 2 , − 1 3 ) . Consider the vertices of square ( 0 , 0 ) , ( 3 , 0 ) , ( 3 , − 3 ) , ( 0 , − 3 ) and critical point ( 2 , − 1 3 ) . The above square is shown below, Figure (1) For segment L 1 , f ( x , 0 ) = x 2 − 4 x − 3 Differentiate above both sides, f ′ ( x ) = 2 x − 4 0 = 2 x − 4 x = 2 So, critical point is ( 2 , 0 ) . For segment L 2 , f ( 0 , y ) = 3 y 2 + 2 y Differentiate above both sides, f ′ ( y ) = 6 y + 2 0 = 6 y + 2 y = − 1 3 So, critical point is ( 0 , − 1 3 ) . For segment L 3 , f ( x , − 3 ) = x 2 − 4 x + 18 Differentiate above both sides, f ′ ( x ) = 2 x − 4 0 = 2 x − 4 x = 2 So, critical point is ( 2 , − 3 ) . For segment L 4 , f ( 3 , y ) = 3 y 2 + 2 y − 6 Differentiate above both sides, f ′ ( y ) = 6 y + 2 0 = 6 y + 2 y = − 1 3 So, critical point is ( 3 , − 1 3 )

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ISBN: 9781524916817

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 11.7, Problem 26PS

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The absolute extrema of f on the closed bounded set.

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The absolute extrema of f on the closed bounded set.

Solution Summary: The author calculates the absolute extrema of f on the closed bounded set.

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