Calcoptimization
posted by Daryl on .
Given y=(x)^1/2, find the closest point to (3/2,0)

let the closest point be P(a,b)
label the given point A(3/2 , 0)
AP will have a minimum length when AP is perpendicular to the tangent at P
slope of tangent = y' = (1/2)x^(1/2) or 1/(2√x)
so at P the slope of the tangent is 1/2√a
slope of AP = b/(a3/2)
b/(a3/2) = 2√a , (the negative reciprocal of 1/2√a
but also b = √a
√a/(a3/2) = 2√a
1/(a3/2) = 2
1 = 2a + 3
2a = 2
a = 1
then b=√1 = 1
the closest point is (1,1)