Optimization Prob.: Find pt on y = x^2 closest to (4, -0.5)

[merin00]

I know it's probably really easy but I have no idea how to start it.

Find the point on the curve y = x^2 that is closest to the point (4, -1/2)

 

[galactus]

\(\displaystyle \L\\S=L^{2}=(x-4)^{2}+(y+\frac{1}{2})^{2}\)

Sub in y=x^2, differentiate, set to 0 and solve for x.

The trick is knowing that the distance and the square of the distance have their max and min at the same point. Thus, we can use \(\displaystyle L^{2}\)