# 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

$$x^2+2y^2=1$$

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

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

Solution

Coming soon…

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