Explanation Given: Function f ( x , y ) = 400 − 3 x 2 − 4 x + 2 x y − 5 y 2 + 48 y Formula Used: Critical points of the function are the points where the function is defined and its derivative is undefined or zero. Mathematically, f x ( x 0 , y 0 ) = 0 f y ( x 0 , y 0 ) = 0 Let us assume that ( x 0 , y 0 ) is the critical point of a continuous function f and D be calculated as follows: D = f x x ( x 0 , y 0 ) f y y ( x 0 , y 0 ) − f x y ( x 0 , y 0 ) 2 Then, If D > 0 and f x x ( x 0 , y 0 ) > 0 , then the function has local minimum at ( x 0 , y 0 ) If D > 0 and f x x ( x 0 , y 0 ) < 0 , then the function has local maximum at ( x 0 , y 0 ) If D < 0 , then the function has saddle point at ( x 0 , y 0 ) If D = 0 , then test is inconclusive Calculation: Function is given as f ( x , y ) = 400 − 3 x 2 − 4 x + 2 x y − 5 y 2 + 48 y Differentiate f ( x , y ) with respect to x , we get f x ( x , y ) = d d x ( 400 − 3 x 2 − 4 x + 2 x y − 5 y 2 + 48 y ) f x ( x , y ) = 0 − 6 x − 4 + 2 y − 0 + 0 f x ( x , y ) = − 6 x + 2 y − 4 Differentiate f ( x , y ) with respect to y , we get f y ( x , y ) = d d y ( 400 − 3 x 2 − 4 x + 2 x y − 5 y 2 + 48 y ) f y ( x , y ) = 0 − 0 − 0 + 2 x − 10 y + 48 f y ( x , y ) = 2 x − 10 y + 48 To calculate the critical point, put f x ( x 0 , y 0 ) = 0 and f y ( x 0 , y 0 ) = 0 , we have − 6 x + 2 y − 4 = 0 − 6 x + 2 y = 4 − 3 x + y = 2

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 8.5, Problem 24P

To determine

To find all critical points for below function and determine each point as local maxima, local minima and saddle point.

f(x,y)=400−3x2−4x+2xy−5y2+48y

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Chapter 8.5, Problem 23P

Chapter 8.5, Problem 25P

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Chapter 8 Solutions

Applied Calculus, 6e WileyPLUS + Loose-leaf

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To find all critical points for below function and determine each point as local maxima , local minima and saddle point. f ( x , y ) = 400 − 3 x 2 − 4 x + 2 x y − 5 y 2 + 48 y

Solution Summary: The author explains the formula used to find all critical points for a function and determine each point as local maxima, local minima and saddle point.

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To find all critical points for below function and determine each point as local maxima, local minima and saddle point.

f(x,y)=400−3x2−4x+2xy−5y2+48y

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Chapter 8 Solutions

To find all critical points for below function and determine each point as local maxima , local minima and saddle point. f ( x , y ) = 400 − 3 x 2 − 4 x + 2 x y − 5 y 2 + 48 y

Solution Summary: The author explains the formula used to find all critical points for a function and determine each point as local maxima, local minima and saddle point.

Videos

To find all critical points for below function and determine each point as local maxima, local minima and saddle point. f(x,y)=400−3x2−4x+2xy−5y2+48y

Want to see the full answer?

Chapter 8 Solutions

To find all critical points for below function and determine each point as local maxima, local minima and saddle point.

f(x,y)=400−3x2−4x+2xy−5y2+48y