Answer :
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 = [tex]D=\sqrt{(7-x-a)^2+7^2}+\sqrt{(a-x)^2+7^2}\\[/tex]
Objective is to minimize the 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 =
[tex]D_{min}=2*\sqrt{(3.5-x)^2+7^2}\\ and\ D_{min}=2*\sqrt{51.25},\ \ \ for\ x=2cm\\[/tex]
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 = [tex]D=\sqrt{(7-x-a)^2+7^2}+\sqrt{(a-x)^2+7^2}\\[/tex]
Objective is to minimize the 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 =
[tex]D_{min}=2*\sqrt{(3.5-x)^2+7^2}\\ and\ D_{min}=2*\sqrt{51.25},\ \ \ for\ x=2cm\\[/tex]
To prove this mathematically, we find the derivative of D wrt a, that is, dD/dx 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.
