Absolute Minimums and Maximums — Question 7

Section 16-4 · Calculus pages 3113–3114

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Question 7

On the unbounded domain ℝ2\mathbb R^2, consider f(x,y)=x2+y2+11+x2+y2.f(x,y)=x^2+y^2+\frac{1}{1+x^2+y^2}.

Tasks

  1. Determine all absolute minima and their value.

  2. Decide whether an absolute maximum exists.

  3. Explain why this example shows that compactness is sufficient, but not necessary, for an absolute minimum to exist.

Question 7 – Solution

Strategy. Reduce the function to the radial variable s=x2+y2≥0s=x^2+y^2\ge 0 and study a one-variable function on an unbounded interval.

Step 1: Radial reduction Let g(s)=s+11+s,s≥0.g(s)=s+\frac 1{1+s},\qquad s\ge 0. Then g′(s)=1−1(1+s)2.g'(s)=1-\frac 1{(1+s)^2}. At s=0s=0, g′(0)=0g'(0)=0, and for every s>0s>0, (1+s)2>1(1+s)^2>1, so g′(s)>0g'(s)>0. Therefore gg increases strictly away from s=0s=0.

It follows that fmin=1 at (0,0).\boxed{f_{\min}=1\text{ at }(0,0)}.

Step 2: No maximum As x2+y2=s→∞x^2+y^2=s\to\infty, g(s)=s+11+s→∞.g(s)=s+\frac 1{1+s}\longrightarrow\infty. Thus ff is unbounded above and has .

Step 3: Interpretation The domain ℝ2\mathbb R^2 is closed but not bounded, so the Extreme Value Theorem does not apply. Nevertheless, direct analysis proves that an absolute minimum is attained. Compactness guarantees both extrema for every continuous function, but a particular continuous function may still attain one or both extrema on a noncompact domain.