Absolute Minimums and Maximums — Question 8

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Question 8

For a real parameter aa, define fa(x,y)=(x−a)2+y2f_a(x,y)=(x-a)^2+y^2 on the square S=[−1,1]×[−1,1]S=[-1,1]\times[-1,1].

Tasks

  1. Determine the absolute minimum value and all minimizing points for every aa.

  2. Determine the absolute maximum value and all maximizing points for every aa.

  3. Treat the tie case a=0a=0 explicitly and interpret the answer as squared distance.

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Question 8 – Solution

Strategy. The function is squared distance from (x,y)(x,y) to (a,0)(a,0); minimize and maximize the two nonnegative coordinate contributions independently.

Step 1: Minimum The y2y^2 term is minimized uniquely at y=0y=0. The closest x∈[−1,1]x\in[-1,1] to aa is the clipped value x*={−1,a<−1,a,−1≤a≤1,1,a>1.x_*= \begin{cases} -1,&a<-1,\\ a,&-1\le a\le 1,\\ 1,&a>1. \end{cases} Hence the unique minimizing point is (x*,0)(x_*,0), and fmin=(max⁡{|a|−1,0})2.\boxed{f_{\min}=\bigl(\max\{|a|-1,0\}\bigr)^2}.

Step 2: Maximum The largest possible y2y^2 is 11, attained at y=±1y=\pm 1. The endpoint of [−1,1][-1,1] farthest from aa is at distance |a|+1|a|+1. Therefore fmax=(|a|+1)2+1.\boxed{f_{\max}=(|a|+1)^2+1}. If a>0a>0, the maximizers are (−1,±1)(-1,\pm 1); if a<0a<0, they are (1,±1)(1,\pm 1).

Step 3: Tie case When a=0a=0, both xx-endpoints are equally far away, so all four corners (±1,±1)(\pm 1,\pm 1) maximize f0f_0 with value 22. The point (0,0)(0,0) uniquely minimizes it with value 00. This agrees exactly with the squared-distance interpretation.

Original worksheet page 2: question and worked solution for 3-4-008

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