Absolute max and min of $f(x, y) = x^{2} + 3y^{2} - y$ over the region $x^{2} + 2y^{2} \leq 1$

Question

Find the absolute maximum and minimum values of the function $$f(x, y) = x^{2} + 3y^{2} - y$$ over the region $x^{2} + 2y^{2} \leq 1$

Answer 1

it fist you must look of the extrema in the inner of the region: $$f(x,y)=x^2+3y^2-y$$ at first we have $$\frac{\partial f(x,y)}{\partial x}=2x$$ $$\frac{\partial f(x,y)}{\partial y}=6y-1$$ and then the extrema on the boundary $$x^2+2y^2=1$$ for the second step consider the function $$g(y)=1-2y^2+3y^2-y$$ you will get the maximimum: $$\frac{3}{2}+\frac{1}{\sqrt{2}}$$ for $$x=0,y=-\frac{1}{\sqrt{2}}$$ and the minimum: $$-\frac{1}{12}$$ for $$x=0,y=\frac{1}{6}$$