How do you find the point on the the graph y=sqrtxy=x which is plosest to the point (4,0)?

1 Answer
May 18, 2015

First note that the distance between an arbitrary point (x,y)=(x,sqrt(x))(x,y)=(x,x) on the graph of y=sqrt(x)y=x and the point (4,0)(4,0) is sqrt((x-4)^2+(sqrt(x)-0)^2)=sqrt(x^{2}-7x+16)(x4)2+(x0)2=x27x+16.

Next, note that the value of xx where d(x)=sqrt(x^{2}-7x+16)d(x)=x27x+16 is minimized is the same value of xx where s(x)=(d(x))^2=x^2-7x+16s(x)=(d(x))2=x27x+16 is minimized.

Since s'(x)=2x-7, s(x) and d(x) are minimized at x=7/2=3.5 (note that s''(x)=2>0 so that the critical point is a minimum).

Therefore, the point on the graph of y=sqrt(x) that is closest to (4,0) is (7/2,sqrt(7/2))\approx (3.5,1.87083). The minimum distance is d(7/2)=sqrt(49/4-49/2+16)=sqrt(15/4)=sqrt{15}/2\approx 1.93649