Question

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

Answer 1

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

Learn more:

if x+y+z=0 then the square of the value of (x+y)^2/xy+(y+z)^2/yz+(z+x)

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Answer 2

Answer:

root 14

Step-by-step explanation:

by coshi scodge inequality method

1. => (x+2y+3z)² is greater than or equal to (x²+y²+z²)(1²+2²+3²)
2. => (x+2y+3z)² is greater than or equal to 1×14
3. therefore (x+2y+3z) = root 14

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