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Global Extremum – Domain of ellipse – Exercise 5392

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

\max f(0,\frac{1}{\sqrt{2}}) \approx 1.4

\min f(0,-\frac{1}{\sqrt{2}}) \approx -1.4

Solution

Coming soon…

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