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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