Question

Find the minimum value of x2+y2+z2 when ax+by+cz=p and prove p2/a2+b2+c2

Answer 1

F(x,y,z)= x² + y² +z²

and the constraint is , g(x,y,z)= a x + b y + c z - p

F'(x,0,0)= 2 x-----(1)

F'(0,y,0)= 2 y------(2)

F'(0,0,y)= 2 z-----(3)

g'(x,0,0)= a-------(4)

g'(0,y,0)= b--------(5)

g'(0,0,z)= c----------(6)

Here F'(x,y,z) and g'(x,y,z) means partial derivative of F and g with respect to x,y and z

F'(x,y,z)= k g'(x,y,z)

Equating (1) and (4), (2) and (5)and (3) and (6)

→2 x=k a, →→ 2 y=k b , →→→2 z =k c

$x=\frac{ka}{2}$,

$y=\frac{kb}{2}$,

$z=\frac{kc}{2}$,

Putting the value of ,x , y and z in , a x + by + c z= p, gives

k(a²+b²+c²)= 2 p

$k=\frac{2 p}{a^2 +b^2+c^2}$

Which gives,

$x=\frac{a p}{a^2 +b^2+c^2}$

$y=\frac{b p}{a^2 +b^2+c^2}$

$z=\frac{c p}{a^2 +b^2+c^2}$

So, minimum value of F(x,y,z) when $x=\frac{a p}{a^2 +b^2+c^2}$

$y=\frac{b p}{a^2 +b^2+c^2}$

$z=\frac{c p}{a^2 +b^2+c^2}$ is

    $=\frac{(a p)^2}{(a^2 +b^2+c^2)^2}+\frac{(b p)^2}{(a^2 +b^2+c^2)^2}+\frac{(c p)^2}{(a^2 +b^2+c^2)^2}\\\\ = p^2\times\frac{a^2+b^2+c^2}{(a^2 +b^2+c^2)^2}\\\\ =\frac{ p^2}{a^2 +b^2+c^2}$