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Let $SNF$ be my path to cross the field. Let $SN = x \; \text{km}$, then $NM =1-x\; \text{km}$. Using Pythogaras' theorem for the right angled triangle $NMF$: $$NF^2 = 1 + (1-x)^2 = x^2 -2x + 2.$$ The total time taken $T$ is given by $$T = {x\over 10} + {(x^2 - 2x + 2)^{1/2}\over 6}$$ By differentiation $${dT\over dx} = {1\over 10} + {2x-2\over 12(x^2 -2x + 2)^{1/2}}.$$ For a maximum or minimum time this derivative is zero, so that |