Calculate the maximum and minimum of function $z = f(x,y) = x^2 - y^2 +2$ subjected to the inequality constraint $D=\{(x,y)| x^2 + \frac{y^2}{4} \leq 1\}$.
My solution:
First form the function $$ g(x,y) = x^2 + \frac{y^2}{4} - c, ~0 \leq c \leq 1. $$ Then form the Lagrangian function $$ L(x,y,\lambda) = x^2 - y^2 + 2 + \lambda\left(x^2 + \frac{y^2}{4} - c\right). $$ Therefore we have $$ \left \{ \begin{array}{ll} L_x' = 2x +2\lambda x = 0\\ L_y' = -2y + \frac{\lambda}{2}y = 0 \\ L_\lambda' = x^2 + \frac{y^2}{4} - c \end{array} \right. $$ After solving above equations, so we can got saddle point like $(\varphi(c),\psi(c))$.
- when $c = 0$, we have $x = y = 0$.
- when $c \neq 0$, there are two kind of solutions: (1 $x = 0$, we have $y = \pm 2\sqrt{c}$; (2 $y=0$, we have $ x = \pm \sqrt{c}$.
My problem is that i cant use second partial derivative test for judging $(0,\pm 2\sqrt{c})$ and $(\pm \sqrt{c},0)$ are maximum or minimum, obviously $AC-B^2 = 0$. How can i do next? Thanks in advance!
Clearly, $\nabla f(x,y)=0\iff(x,y)=(0,0)$, but $(0,0)$ is a saddle-point. So, the maximum and the minimum can be attained only at the boundary of $D$.
Let $h(x,y)=x^2+\frac{y^2}4$. Then apply the method of Lagrange multipliers:$$\left\{\begin{array}{l}f_x(x,y)=\lambda h_x(x,y)\\ f_y(x,y)=\lambda h_y(x,y)\\h(x,y)=1,\end{array}\right.$$or$$\left\{\begin{array}{l}2x=2\lambda x\\-2y=\frac12\lambda y\\x^2+\frac{y^2}4=1.\end{array}\right.$$It's solutions are $(x,y,\lambda)=(0,-2,-4)$, $(x,y,\lambda)=(0,2,-4)$, $(x,y,\lambda)=(-1,0,1)$ and $(x,y,\lambda)=(1,0,1)$. Since $f(0,\pm2)=-2$, and $f(\pm1,0)=3$, the maximum is $3$ and the minimum is $-2$.