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

 Apr 3, 2024
 #1
avatar+128794 
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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 + (8)^2 + 12 (-6) -16(8) + 8   =   -92

 

 

cool cool cool

 Apr 4, 2024

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