Let x and y be real numbers such that \(x^2 - xy + y^2 = 3. \)
Let m be the minimum value of \(x^2 + y^2,\) and let M be the maximum value of \(x^2 + y^2.\)Find m+M
+6248 \(\text{There are a few ways to go about this. Lagrange multipliers is one}\\ f = x^2 + y^2\\ g = (x^2 - xy+y^2 - 3)\\ \text{Solve }\nabla(f+\lambda g)=0\)
\(\text{This results in 3 equations}\\ 2x + \lambda(2x-y)=0\\ 2y+\lambda(2y-x) = 0\\ x^2 - xy +y^2 = 3\)
\(\text{This system has 4 solutions}\\ (x,y) = \pm(1,-1)\\ (x,y) = \pm (\sqrt{3},\sqrt{3})\\ \text{resulting in min and max of }x^2 + y^2 \text{ being }2 \text{ and } 6\\ m+M=2+6 = 8\)
I leave the gritty details to you.
