Application of differentiation

y = x^2/2, d^2 = (x + 6)^2 + y^2 and we want to maximize it

Sub. y = x^2/2 into d^2

(x + 6)^2 + y^2 = (x + 6)^2 + x^4/4

Call it f(x), the f'(x) = x^3 + 2x + 12 = (x + 2)(x^2 - 2x + 6)

There is only one real solution x = -2 and so y = 2.

By physical consideration on second derivative test, (-2,2) is a minimum point with the shortest distance √[(-2 + 6)^2 + 2^2] = √20


2011-01-19 21:17:43 補充：
"on" should be "or"