The 3-D Coordinate System — Question 5

Section 14-1 · Calculus pages 2649–2650

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Question 5

A rectangular laboratory is aligned with the coordinate axes and has opposite corners O=(0,0,0),V=(8,6,4).O=(0,0,0),\qquad V=(8,6,4). A wireless transmitter will be installed at a point PP inside the laboratory. Define R(P)=max⁡{d(P,C):C is one of the eight corners}.R(P)=\max\{d(P,C): C\text{ is one of the eight corners}\}.

Design objective

  1. Find the location of PP that minimizes R(P)R(P).

  2. Find the minimum possible value of R(P)R(P).

  3. Prove that no other transmitter location can produce a smaller greatest distance.

Question 5 – Solution

Strategy Opposite corners force a lower bound: for any transmitter location, at least one of its distances to an opposite pair must be at least half the pair’s separation. The box center attains that bound for every pair.

Diagram shown in the original SVG worksheet.

Lower bound The space diagonal has length d(O,V)=82+62+42=116=229.d(O,V)=\sqrt{8^2+6^2+4^2}=\sqrt{116}=2\sqrt{29}. For any location PP, the triangle inequality gives d(O,V)≤d(O,P)+d(P,V).d(O,V)\le d(O,P)+d(P,V). Therefore at least one of d(O,P)d(O,P) and d(P,V)d(P,V) is at least 29\sqrt{29}.

Attaining the bound The midpoint of every space diagonal is M=(4,3,2).M=\left(4,3,2\right). Each corner differs from MM by coordinates ±4,±3,±2\pm 4,\pm 3,\pm 2, so every corner is exactly 42+32+22=29\sqrt{4^2+3^2+2^2}=\sqrt{29} units away. Hence the optimal location is (4,3,2)\boxed{(4,3,2)}, and the minimized greatest distance is 29\boxed{\sqrt{29}}.

Verification The lower bound and attained value agree, which proves global optimality rather than merely checking a plausible point.