STEM-Master
Calculus IIChapter 6: 3-Dimensional SpaceSection 6-5

Functions of Several Variables — Question 10

PDF ↗

Question 10

Let f(x,y)=xy1+x2+y2f(x,y)=\dfrac{xy}{1+x^2+y^2} on the closed rectangle R=[−2,2]×[−3,3]R=[-2,2]\times[-3,3]. Prove that |f|<1/2|f|<1/2 everywhere, and explain why ff must attain a maximum and minimum on RR even without finding them.

Original worksheet page 1: question and worked solution for 6-5-010
Show solutionHide solution

Question 10 – Solution

Strategy Compare 2|xy|2|xy| with x2+y2x^2+y^2, then use continuity on a closed bounded domain.

See the diagram in the original worksheet below.

Bound From (|x|−|y|)2≥0(|x|-|y|)^2\ge 0, 2|xy|≤x2+y22|xy|\le x^2+y^2. Hence |f(x,y)|≤x2+y22(1+x2+y2)<12.|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 ff is continuous. The rectangle RR is closed and bounded (compact). By the Extreme Value Theorem, ff attains both an absolute maximum and absolute minimum on RR.

Original worksheet page 2: question and worked solution for 6-5-010

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

← Section questionsChapter contents