Find the points on the plane x + y + z = 9 which are closest to origin
Answers
Given : plane x + y + z = 9
To Find : points on the plane x + y + z = 9 which are closest to origin
Solution:
x + y + z = 9
=> x = 9 - (x + y)
Points are ( x , y , 9 - (x + y))
Distance from Origin d= √x² + y² + (9 - ( x+ y))²
z = d²
Hence Z = x² + y² + (9 - ( x+ y))²
=> Z = x² + y² + 81 + x² + y² + 2xy - 18x - 18y
=> Z = 2x² + 2y² + 2xy - 18x - 18y + 81
∂Z/∂x = 4x + 2y - 18
∂Z/∂y = 4y + 2x - 18
∂Z/∂x = 0 and ∂Z/∂y = 0
=> 4x + 2y - 18 = 0 and 4y + 2x - 18 = 0
on solving x = 3 , y = 3
=> ( 3 , 3)
∂²Z/∂x² = 4 > 0
∂²Z/∂y² = 4 > 0
Minimum Distance when
x = 3 , y = 3
z = 9 -(x + y)
=> z = 3
points on the plane x + y + z = 9 which are closest to origin = 3 , 3 ,3
3 , 3 ,3 is the point on the plane x + y + z = 9 which are closest to origin
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