Functions of Several Variables — Question 10

Question 10

Let $f(x,y)=\dfrac{xy}{1+x^2+y^2}$ on the closed rectangle $R=[-2,2]\times[-3,3]$. Prove that $|f|<1/2$ everywhere, and explain why $f$ must attain a maximum and minimum on $R$ even without finding them.

Question 10 – Solution

Strategy Compare $2|xy|$ with $x^2+y^2$, then use continuity on a closed bounded domain.

Bound From $(|x|-|y|)^2\ge 0$, $2|xy|\le x^2+y^2$. Hence $$|f(x,y)|\le\frac{x^2+y^2}{2(1+x^2+y^2)}<\boxed{\frac 12}.$$ The strict inequality comes from the added 1 in the denominator.

Existence of extrema The denominator never vanishes, so $f$ is continuous. The rectangle $R$ is closed and bounded (compact). By the Extreme Value Theorem, $f$ attains both an absolute maximum and absolute minimum on $R$.