Applied optimization
Find the point on the curve y=√(2x+2) which is closest to the point (4,0).
SOLUTION
This is a distance problem with the goal of minimizing the distance between a point on the curve y=√(2x+2) and the point (4,0).
Let's use the distance formula d = √[(x-x1)^2 + (y-y1)^2]
Substituting values for a point on the curve and (4,0) gives us
d(x) = √[(x-4)^2 + (√(2x+2)-0)^2]
d(x) = √[(x^2 - 8x +16) + (2x + 2)]
d(x) = √[x^2 - 6x +18]
Take the first derivative of d(x) to find the minimum distance.
d'(x) = (1/2)[(x^2 - 6x +18)^(-1/2)](2x - 6)
d'(x) = [(x^2 - 6x +18)^(-1/2)](x - 3)
Set d'(x) = 0 to find the value of x at the minimum distance.
d'(x) = 0 when the numerator x - 3 = 0.
So, solving x - 3 = 0
Results in x = 3 and y = √(2·3+2) = √8 = 2√2 for the point (3, 2√2) and the minimum distance is d(3) = 3.
