Extremum value

Algebra Level 4

If x 2 + y 2 + z 2 = 1 x^2+y^2+z^2=1 , then what is maximum value of ( 2 z − 3 y ) 2 + ( 3 x − z ) 2 + ( y − 2 x ) 2 ? (2z-3y)^2+(3x-z)^2+(y-2x)^2?


The answer is 14.

Sameer Mishra by Sameer Mishra , 31, India

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2 solutions

Dan Huang
Aug 31, 2019

Here is a solution without vectors.

( x + 2 y + 3 z ) 2 + ( 2 z − 3 y ) 2 + ( 3 x − z ) 2 + ( y − 2 x ) 2 = 14 ( x 2 + y 2 + z 2 ) = 14 (x+2y+3z)^2 + (2z-3y)^2+ (3x-z)^2 + (y-2x)^2 = 14(x^2+y^2+z^2)=14

( 2 z − 3 y ) 2 + ( 3 x − z ) 2 + ( y − 2 x ) 2 = 14 − ( x + 2 y + 3 z ) 2 (2z-3y)^2+ (3x-z)^2 + (y-2x)^2 = 14 - (x+2y+3z)^2

Therefore, in order to maximize the left side of the equation, we have to minimize ( x + 2 y + 3 z ) 2 (x+2y+3z)^2 . Since it is a square, the minimum value is 0. The values ( 1 / 3 , 1 / 3 , − 1 / 3 (\sqrt{1/3} , \sqrt{1/3}, -\sqrt{1/3} ) work.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Sameer Mishra
Mar 2, 2018

Let vector A=xi+yj+zk,B=1i+2j+3k now take magnitude of A×B then get 14^1/2×(x^2+y^2+z^2)sinp,but max(sinp)=1and x^2+y^2+z^2=1 so maximum value is 14

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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