Find the maximum and minimum value of $f(x,y)=2x^2+3y^2$ on $x^2+y^2 \leq 1$

Question

$f(x,y)=2x^2+3y^2$ , $g(x,y)=x^2+y^2 \leq 1$

Let $G(x,y)=x^2+y^2-1=0$

Using lagrange multiplier

$f_{x}(x,y)=4x$ , $f_{y}(x,y)=6y$

$G_{x}(x,y)=2x$ , $G_{y}(x,y)=2y$

$f_{x}(x,y)= \lambda G_{x}(x,y)$

$f_{y}(x,y)= \lambda G_{y}(x,y)$

$x(2-\lambda)=0 $

$y(3-\lambda)=0 $

Either $x=0$ or $\lambda = 2$

If $x=0$, $y=1$ or $y=-1$ from $G(x,y)$

If $\lambda=2$ then $y(3-2)=0 $ , $y$=0

If $y=0$, $x=1$ or $x=-1$ from $G(x,y)$

The max value of $f(x,y)$ is $f(0,1)=f(0,-1)=3 $

The min value of $f(x,y)$ is $f(1,0)=f(-1,0)=2$

Somehow the minimum value is incorrect here but I don't understand why

Answer 1

Here's a slightly different approach. First of all, we are going to find extreme points of $f$, if any, in the disk $D=\{(x,y): x^2+y^2<1\}$ . Then, we are going to find extreme points of $f$, if any, on the boundary of the disk $D$, i.e. on $C=\{(x,y): x^2+y^2=1\}$.

Since $$ f_x(x,y)=4x,f_y(x,y)=6y, $$ we have $f_x(x,y)=0$ amd $f_y(x,y)=0$ for $(x,y)=(0,0)$.

Because $(0,0)$ is inside $D$ and $f(x,y)\ge 0=f(0.0)$ for all $(x,y)$ in $\mathbb{R}^2$, it is clear that $f(0,0)=0$ is the absolute minimum of $f$.

Now, let's examine $f$ on $C$. Using the parametrization $$ x=\cos t, y=\sin t, \quad 0 \le t <2\pi, $$ we can represent $f_{|C}$ as $$ \varphi(t)=2\cos^2t+3\sin^2t=2+\sin^2t,\quad 0\le t <2\pi. $$

We have \begin{eqnarray}\tag{1} \min\varphi &=&\varphi(0)=\varphi(\pi)=2\\ \max\varphi &=& \varphi(\pi/2)=\varphi(3\pi/2)=3 \end{eqnarray}

Thus, the absolute maximum of $f$ on $\bar{D}=\{(x,y): x^2+y^2\le 1\}$ is $$ f(0,\pm1)=3. $$

Notice that the absolute maximum of $f$ on $\bar{D}$ lies on $C$. Therefore, one can also use Lagrange Multipliers to find the maximum of $f$ (on $C$). In order to do so, we consider the function $$ F(x,y,\lambda)= 2x^2+3y^2+\lambda(x^2+y^2-1) $$ At a critical point we'll have \begin{eqnarray}\tag{1} 0 &=& F_x(x,y,\lambda)=2x(2+\lambfa)\\ 0 &=& F_y(x,y,\lambda)=2y(3+\lambfa)\\ 0 &=&F_{\lambda}(x,y,\lambda)=x^2+y^2-1 \end{eqnarray} Splving the first equation of system (1), we have $x=0$ or $\lambda=-2$.

If $x=0$, the last equation of system (1) yields $y=pm1$.

If $\lambda=-2$, the second equation of system (1) is equivalent to $y=0$, and the last equation of (1) then gives us $x=\pm1$. Hence, the first equation of (1) gives the points $$ (0,\pm1), (\pm1,0) $$ Similarly, solvung the second equation of (1) hives the same points.

Since $$ f(0.0)=0< f(\pm1,0)=2 <f(0,\pm1)=3, $$ the minimum and maximum of $f$ on $\bar{D}$ are $f(0.0)=0$ and $f(0,\pm1)=3$, respectively.

Answer 2

In short, your minimum is incorrect because you are finding the extrema of $f$ given the condition $x^2 + y^2 = 1$; that is to say, you are only looking at the boundary of the unit disk, rather than all points inside the disk.

The minimum on the disk $x^2 + y^2 \le 1$ is simply $0$, attained at $(x,y) = (0,0)$. I leave it to you as an exercise to set up the appropriate equations to show this via Lagrange multipliers.