Global Extremum – Domain of a circle – Exercise 6543 Post category:Global Extremum Post comments:0 Comments Exercise Find the maximum value and the minimum value of the function z(x,y)=4x^2-4xy+y^2 In the domain: D=\{ (x,y):x^2+y^2=25\} Final Answer Show final answer \max_D z(-2\sqrt{5},\sqrt{5})=\max_D z(2\sqrt{5},-\sqrt{5})= 125 \min_D z(\sqrt{5},2\sqrt{5})=\min_D z(-\sqrt{5},-2\sqrt{5})=0 Solution Coming soon… Share with Friends Read more articles Previous PostGlobal Extremum – Domain of a circle – Exercise 6538 Next PostGlobal Extremum – Domain of a function with fixed negative powers – Exercise 6551 You Might Also Like Global Extremum – Domain of lines – Exercise 5529 June 9, 2019 Global Extremum – Domain of ellipse – Exercise 4749 May 6, 2019 Global Extremum – Domain of a circle – Exercise 6538 July 15, 2019 Global Extremum – Domain of ellipse – Exercise 5392 May 15, 2019 Global Extremum – Domain of a function with fixed negative powers – Exercise 6551 July 23, 2019 Global Extremum – Domain of lines – Exercise 3443 February 21, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ