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

d^{2} = s^{2} + (s - x)^{2} = x^{2} + 2 s^{2} - 2 sx

where

s

is the length of the edge of the field. The time taken

t

is given by

t = \frac{x}{10} + \frac{(x^{2} + 2 s^{2} - 2 sx)^{1 / 2}}{6}

By differentiation

\frac{dt}{dx} = \frac{1}{10} + \frac{2 x - 2 s}{12 (x^{2} + 2 s^{2} - 2 sx)^{1 / 2}}

. For

t

to be a minimum

dt / dx = 0

and

3 (x^{2} + 2 s^{2} - 2 sx)^{1 / 2} = 5 (s - x)

so that

9 (x^{2} + 2 s^{2} - 2 sx) = 25 (s^{2} - 2 sx + x^{2})

which gives

16 x^{2} - 32 sx + 7 s^{2} = 0

which factorizes to give

(4 x - s) (4 x - 7 s) = 0

x = s / 4

x = 7 s / 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.

$t = \frac{1 - \cot θ}{10} + \frac{1}{6 \sin θ} .$