Question

Exercise: An ant lives on the surface of a cube with edges of length 7cm. It is currently
located on an edge x cm from one of its ends. While traveling on the surface of the cube,
it has to reach the grain located on the opposite edge (also at a distance xcm from one
of its ends) as shown below.
(i) What is the length of the shortest route to the grain if x = 2cm? How many routes of
this length are there?
(ii) Find an x for which there are four distinct shortest length routes to the grain.
Scope and Effort
Total expected

Answer 1

see diagram.

The ant is at P initially.  It goes to R via Q. Let the distance of Q from left corner be equal to a.

distance total = PQ + QR =
$D=\sqrt{(7-x-a)^2+7^2}+\sqrt{(a-x)^2+7^2}\\\\ Objective\ is\ to\ minimize\ that\ sum\ D.$
Both terms are positive.  Both are symmetrically placed on the cube.  From symmetry we can infer that the total is minimum when both terms are equal.  We can mathematically prove and show that.

     So, 7 - x - a = a - x    =>  a  = 3.5.    => Q is in the center of the edge.

The shortest distance will be D =
$D_{min}=2*\sqrt{(3.5-x)^2+7^2}\\\\D_min=2*\sqrt{51.25},\ \ \ for\ x=2cm$

To prove this mathematically,  we find the derivative of D wrt a, that is, $\frac{dD}{da}$  and then equate it to 0.  Then we get an equation in a.  Solving and taking the positive value, we get the value of a (less than 7, between x and 7-x)  for which D is minimum. 
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  There will be two routes in general for x ≠ 7/2cm.  The two routes will be on the opposite faces of the cube.  In the diagram shown, the other route will be on the faces UTS and WVS.

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ii)   If the ant P is at one vertex of the cube, then there will be 4 distinct equal routes to the diagonally opposite vertex of the cube.  The route will be along an edge (7cm) + a diagonal on the face (7 * root (2) ) .  That will be the total distance.