I want to find the minimum and maximum values of the function $f(x, y) = \ln(1 + 9x^2 + 4y^2)$ within the region $1 \leq \frac{x^2}{4} + \frac{y^2}{9} \leq 4$. My solution steps are as follows:
- Calculated the partial derivatives of $f$ with respect to $x$ and $y$.
- Found the critical points by solving the system of equations $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
- Evaluated the function at the critical point (0, 0)
I obtained the following result:
- Absolute Minimum: 0 (at the critical point (0, 0)).
Could you please confirm if my solution is correct or if there is any error in my approach? I appreciate your help in verifying my work.
The region $$D=\Big \{(x,y):1 \leq \frac{x^2}{4} + \frac{y^2}{9} \leq 4\Big\}$$ is a compact set and therefore the differentiable function $f$ has minimum and maximum values in $D$.
You are right, $f$ has a unique critical point at $(0,0)$, but $(0,0)\not\in D$ This fact implies that the minimum points and the maximum points stay on the boundary of $D$ which is the union of two ellipses $E_1$ and $E_4$ with $$E_r=\Big \{(x,y):\frac{x^2}{4} + \frac{y^2}{9} =r\Big\}.$$
Notice that along any ellipse $E_r$ with $r\geq 0$, the function $f$ is constant: $$\ln(1 + 9x^2 + 4y^2)= \ln\Big(1 + 36\Big(\frac{x^2}{4} + \frac{y^2}{9}\Big)\Big)=\ln(1 + 36r)$$ which is increasing with respect to $r$. Hence the minimum value of $f$ is attained along $E_1$ where $f$ is identically equal to $\ln(37)$. Similarly, the maximum value of $f$ is attained along $E_4$ where $f$ is identically equal to $\ln(145)$.