Expert: Steve Holleran - 10/17/2007

Question
I actually have two questions:

1.Find two numbers whose product is -16 and the sum of whose squares is a minimum.

2. Find the points on the parabola x=2y^2 that are closest to the point (10, 0).


Answer
Hi Alex,

1.  Okay, here, we have xy = -16 , so ly = -16/x or -16x^-1

Then we want to minimize

S = x^2 + y^2 = x^2 + (-16/x)^2 = x^2 + (256/x^2)

 = x^2 + 256x^-2

So S' = 2x - 512x^-3 = 2x -(512/x^3) = 0

when 2x = 512/x^3  so x^4 = 512/2 = 256 and x = 4.

Also, S'' = 2 + 1526/x^4 and S''(4) is positive, so we have a minimum CV.  So, x = 4 and y = -4.



2.  Here we want to minimize the distance from a point on the parabola, (x,y) to (10,0)

since x = 2y^2, call the point (2y^2,y).

Then the distance formula is

d = sqrt[ (2y^2 - 10)^2 + (y-0)^2]

Call D = d^2 = (2y^2-10)^2 + y^2.  Now, d will be minimized when D is minimized so

D' = 2(2y^2-10)(4y) + 2y = 8y(2y^2-10) + 2y

  = 16y^3 - 80y +2y = 16y^3 -78y

  = 2y(8y^2 - 39) = 0  when y = 0 and y = sqrt(39/8)

Then D'' = 48y^2 - 78 and this is positive if y = sqrt(39/8)

so this is a minimum C.V.

Then

x = 2 * 39/8 = 39/4

and the point is (39/4 , sqrt(39/8)).


Hope this helps you out.

Steve