Answer
$$global~max: 1,\ \ \ \ \ global~min: \frac{1}{35}$$
Work Step by Step
Given $$f(x, y)=\left(x^{2}+y^{2}+1\right)^{-1}, \quad 0 \leq x \leq 3, \quad 0 \leq y \leq 5$$
We see that $f(x,y)$ is maximum when $x^2$ and $y^2$ are minimum
(when $x = y = 0$). $f$ is minimum when $x^2$ and $y^2$ are maximum; that is, when $x = 3$ and $y = 5$.
Hence, the global maximum of $f$ on the given set is
$$f (0, 0) = (02 + 02 + 1)^{-1} = 1$$ and the global minimum is $$f (3, 5) = (3^2 + 5^2 + 1)^{-1} =\frac{1}{35} $$
