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What is the minimum value of the expression $x^2+y^2-6x+4y+18$ for real $x$ and $y$?

 Nov 3, 2014

Best Answer 

 #1
avatar+23254 
+7

x² + y² - 6x + 4y + 18

Rearrange:

x² - 6x + y² + 4y + 18

Complete the squares:

(x² - 6x   ) + (y² + 4y   ) + 18

(x² - 6x + 9) + (y² + 4y + 4) + 18 - 9 - 4

(x - 3)² + (y + 2)² + 5

Since (x - 3)² can never be less than 0, and neither can (y + 2)², the complete expression must always be at least 5, but can be larger than 5.

 Nov 3, 2014
 #1
avatar+23254 
+7
Best Answer

x² + y² - 6x + 4y + 18

Rearrange:

x² - 6x + y² + 4y + 18

Complete the squares:

(x² - 6x   ) + (y² + 4y   ) + 18

(x² - 6x + 9) + (y² + 4y + 4) + 18 - 9 - 4

(x - 3)² + (y + 2)² + 5

Since (x - 3)² can never be less than 0, and neither can (y + 2)², the complete expression must always be at least 5, but can be larger than 5.

geno3141 Nov 3, 2014

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