I got the function $f(x,y)=\ln(1+x^2+y^2)$. There are three tasks to answer.
a)Decide the function´s stationary points and classify them if possible. Here I got the answer to $(0,0)$ is a local maximum point.
b)Decide a Here I got the answer $(0,0)$
c)Now limit the domain of definition to $x^2+y^2≤1$. Decide the function´s biggest and lowest value and the range.
Please help me with task c). The answer should be: lowest value: $f(0,0)=0$. Biggest value: $f(a,b)=\ln(2)$ for all $a^2+b^2=1$. The range is $0≤z≤\ln(2)$. How do I get there?
$f(x,y)=\ln(1+x^2+y^2)$
Note that the function $\ln(1+z)$ is increasing with respect to $z$. Therefore $\ln(1+x^2+y^2)$ can be maximised or minimised be maximising or minimising $x^2+y^2$, respectively. $x^2+y^2=1$ is the maximum value, and has infinitely many solutions, so the maximum of $f(x,y)=\ln(1+1)=\ln2$ whenever $x^2+y^2=1$. $x^2+y^2=0$ is the minimum, and occurs only when $x=0,y=0$. Therefore $f(x,y)$ is minimised at $f(0,0)=\ln1=0.$
This is in the same vein as my answer to this question.