Question

If the value of x^2+y^2+z^2=1 .then what is the maximum value of x+2y+3z

Answer 1

3xyz
hope it helps you

Answer 2

Given :   X^2 + y^2 +z^2 =1

To find :     maximum value of x+2y +3z?

Solution:

x + 2y + 3z can be represented as dot product  of two vectors

x + 2y + 3z  = ( xi + yj + zk)  . ( i + 2j + 3k )

as we know

u . v = |u | | v| Cosθ

( xi + yj + zk)  . ( i + 2j + 3k ) = |√x² + y² + z² |. |√1² + 2² + 3² | Cosθ

=> x + 2y + 3z  = 1 * √14 Cosθ

Maximum value of  Cosθ = 1

Hence maximum value of x + 2y + 3z =  √14

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