Global Extremum – Domain of a function with fixed negative powers – Exercise 6551 Post category:Global Extremum Post comments:0 Comments Exercise Find the maximum value and the minimum value of the function z(x,y)=1+1x+1yz(x,y)=1+\frac{1}{x}+\frac{1}{y}z(x,y)=1+x1+y1 In the domain: D={(x,y):1x2+1y2=18}D=\{ (x,y):\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{8}\}D={(x,y):x21+y21=81} Final Answer Show final answer maxDz(−25,5)=maxDz(25,−5)=125\max_D z(-2\sqrt{5},\sqrt{5})=\max_D z(2\sqrt{5},-\sqrt{5})= 125Dmaxz(−25,5)=Dmaxz(25,−5)=125 minDz(5,25)=minDz(−5,−25)=0\min_D z(\sqrt{5},2\sqrt{5})=\min_D z(-\sqrt{5},-2\sqrt{5})=0Dminz(5,25)=Dminz(−5,−25)=0 Solution Coming soon… Share with Friends Read more articles Previous PostGlobal Extremum – Domain of a circle – Exercise 6543 You Might Also Like Global Extremum – Domain of ellipse – Exercise 5392 May 15, 2019 Global Extremum – Domain of a circle – Exercise 6538 July 15, 2019 Global Extremum – Domain of lines – Exercise 5529 June 9, 2019 Global Extremum – Domain of a circle – Exercise 6543 July 15, 2019 Global Extremum – Domain of ellipse – Exercise 4749 May 6, 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