Let's do the unit cube and worry about the side length later.
Let's assign F(0,0,0), H(1,1,0), A(0,1,1), C(1,0,1)
Let's call the beetle positions P and Q, functions of time t.
P = F + 2t(H-F) = tH = 2t(1,1,0) = (2t,2t,0)
Q = A + t(C-A) = (0,1,1)+t(1,-1,0)=(t, 1-t, 1)
The squared distance is
[tex]PQ^2 = (2t - t)^2 + (2t -(1-t))^2 + 1 = t^2 + (3t-1)^2 + 1 = 10t^2-6t+2[/tex]
[tex]\dfrac{d\ PQ^2}{dt} = 20t - 6 = 0[/tex]
[tex]t = 6/20 = 3/10 [/tex]
[tex]PQ^2 = 10(3/10)^2 - 6(3/10) + 2 = 11/10[/tex]
[tex]\sqrt{11/10} \times 40 \sqrt{110} = 440 [/tex]
Answer: 440
