If you go along the edge a distance x then head straight for the far corner the distance travelled across the field d is given by
d2 = s2 + (s-x)2 = x2 + 2s2 -2sx
where s is the length of the edge of the field. The time taken t is given by
t = x
10
 + (x2 + 2s2 -2sx)1/2
6
By differentiation
dt
dx
 = 1
10
 + 2x-2s
12(x2 + 2s2-2sx)1/2
. For t to be a minimum dt/dx = 0 and
3(x2 + 2s2-2sx)1/2 = 5(s-x)
so that
9(x2 + 2s2-2sx) = 25(s2 -2sx +x2)
which gives
16x2 -32sx +7s2 = 0
which factorizes to give
(4x-s)(4x-7s) = 0
so x=s/4 or x=7s/4.

Alternatively t can be expressed in terms of the angle taken across the field (taking s=1) and then you can differentiate wrt the angle.
t = 1-cotq
10
 + 1
6sinq
 .