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 $\displaystyle L^{2}$