Global Extremum – Domain of ellipse – Exercise 5392 Post category:Global Extremum Post comments:0 Comments Exercise Use the Lagrange multiplier method to find a maximum value and a minimum value of the function f(x,y)=x^2+4y^3 On the ellipse x^2+2y^2=1 Final Answer Show final answer \max f(0,\frac{1}{\sqrt{2}}) \approx 1.4 \min f(0,-\frac{1}{\sqrt{2}}) \approx -1.4 Solution Coming soon… Share with Friends Read more articles Previous PostGlobal Extremum – Domain of ellipse – Exercise 4749 Next PostGlobal Extremum – Domain of lines – Exercise 5529 You Might Also Like Global Extremum – Domain of ellipse – Exercise 4749 May 6, 2019 Global Extremum – Domain of a function with fixed negative powers – Exercise 6551 July 23, 2019 Global Extremum – Domain of a circle – Exercise 6538 July 15, 2019 Global Extremum – Domain of a circle – Exercise 6543 July 15, 2019 Global Extremum – Domain of lines – Exercise 5529 June 9, 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