Question

x^2+y^2=1 then what is the maximum value of xy​

Answer 1

Step-by-step explanation:

Given X^2+y^2+z^2 =1

To find :

maximum value of x+2y +3z?

3

3.0

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=lu || vl Cose

(xi+yj + zk). (i+2j+3k) = √x² + y² + z²1. |√1² +2²+3² | Cose

=> x + 2y + 3z =1* √14 Cose Maximum value of Cose = 1

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

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