Express the square of the distance between an arbitrary point (x, y) on the curve and (1, 0). It is easy to find the minimum of this expression.

The fireman would do well to use whatever ladder he has instead of doing calculus. Let the base of the ladder be x meters from the building wall and let the top of the ladder touch the wall at y meters above the ground. Assuming the ladder touches the top of the fence, from the similarity of triangles we have y / x = 2 / (x - 1). Express y through x, find the square of the ladder length y^{2}+ x^{2}in terms of x, then find the minimum of this expression using differentiation.