Find the minimum distance from the point 1 1 14 to the parab

Find the minimum distance from the point (1, 1, 14) to the paraboloid given by the equation z = x^2 + y^2. Minimum distance = 0.2184319

Solution

point on paraboloid z=x²+y² is (x,y,z)

distance from point (1,1,14) to paraboloid is , d=[(x-1)²+(y-1)²+(z-14)²]

distance is minimum if f(x,y,z)=(x-1)²+(y-1)²+(z-14)² is minimum

we have z=x²+y²

so f(x,y)=(x-1)²+(y-1)²+(x²+y²-14)²

f_x =(2(x-1))+0+(2(x²+y²-14)*2x), f_y =0+(2(y-1))+(2(x²+y²-14)*2y)

f_x =(2(x-1))+(4x(x²+y²-14)), f_y =(2(y-1))+(4y(x²+y²-14))

for critical points f_x =0, f_y =0

(2(x-1))+(4x(x²+y²-14))=0, (2(y-1))+(4y(x²+y²-14))=0

(x-1)+(2x(x²+y²-14))=0, (y-1)+(2y(x²+y²-14))=0

(x²+y²-14)=(1-x)_/2x ,(x²+y²-14)=(1-y)_/2y

(1-x)_/2x =(1-y)_/2y

=>x=y

(x-1)+(2x(x²+y²-14))=0

=>(x-1)+(2x(x²+x²-14))=0

=>x-1+4x³-28x =0

=>4x³-27x -1=0

=>x=-2.5794,-0.037045,2.6164

x=y

=>y=-2.5794,-0.037045,2.6164

z=x²+y²

=>z=13.30661, 0.002744664 ,13.6911

at (-2.5794,-2.5794,13.30661) distance d=[(-2.5794-1)²+(-2.5794-1)²+(,13.30661-14)²]=5.10931

at (-0.037045,-0.037045, 0.002744664) distance d=[(-0.037045-1)²+(-0.037045-1)²+(0.002744664-14)²]=14.0739

at (2.6164,2.6164,13.6911) distance d=[(2.6164-1)²+(2.6164-1)²+(13.6911-14)²]=2.30671

minimum distance =2.30671