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What is the minimum value of the expression \(2x^2+3y^2+8x-24y+62\) for real x and y?

 Aug 24, 2018
 #1
avatar+101796 
+2

f (x, y)  = 2x^2  + 3y^2 + 8x - 24y +62

 

Take partial derivatives in x and y and set these to 0

 

fx   = 4x + 8  = 0    →  x  = -2

fy = 6y - 24  = 0     → y  = 4

 

As WolframAlpha shows....(-2,4)  is the min for this function.....and the min value  = 6

 

https://www.wolframalpha.com/input/?i=minimize+%5B+2x%5E2%C2%A0+%2B+3y%5E2+%2B+8x+-+24y+%2B+62%5D

 

 

cool cool cool

 Aug 25, 2018
 #2
avatar+814 
+2

Thanks! Very good!

mathtoo  Aug 25, 2018

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