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Find the minimum value of the expression $x^2+y^2+2x-4y+8+10x-12y$ for real $x$ and $y$.

 Oct 12, 2023
 #1
avatar+128826 
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Simplify as

 

x^2 + 12x + y^2 - 16y + 8

 

Take the derivative with respect to x  and set  to 0

2x + 12 = 0

x = -6

 

Take the derivative with respect to y and  set  to  0

2y - 16  = 0

y = 8

 

Minimum  is

 

(-6)^2  + 12 (-6) + (8)^2 - 16 (8)  + 8  =   -92

 

 

cool cool cool

 Oct 12, 2023

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