find the point on f(x)=x^2 in the first quadrant that is closest to the point (0, 9/2)

Question

find the point on f(x)=x^2 in the first quadrant that is closest to the point (0, 9/2)

Answer 1

The distance from any point (x,y) to the point (0,9/2) is given by the distance formula  

 

d = sqrt[ (x - 0)^2 + (y - 9/2)^2]  

 

d will be minimized when g = d^2 is minimized 

 

g = x^2 + (y - 9/2)^2 

 

If the point (x,y) lies on f(x) = x^2 then y = x^2 so 

 

g = x^2 + (x^2 - 9/2)^2 = x^2 + x^4 -9*x^2 + 81/4 = x^4 - 8*x^2 + 81/4  

 

g'(x) = 4*x^3 - 16*x = 4*x*(x - 2)*(x + 2)  

 

g'(x) = 0 when x = -2, 0 or 2.  Only 2 is in the first quadrant.  

 

g''(x) = 12x^2 - 16

g''(2) = 48 - 16 = 32 > 0 so x = 2 is a minimum 

 

So (2,4) is the point on f(x) = x^2 in the first quadrant that is closest to the point (0,9/2)