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Among all pairs of numbers (x,y) such that 3x+y=15, find the minimum of x^2 + y^2.

 
 Nov 21, 2020
 #1
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We  are looking for the  minimum (squared) distance  from the origin to the line   3x + y  =15

 

Let this point  be   (x  , 15 - 3x )

 

D^2  =  ( x- 0)^2  +  ( 15  -3x-0) ^2

 

D^2  =  x^2  + ( 15-3x)^2

 

D^2  = x^2  + 225 - 90x + 9x^2      take the derivative and set to 0

 

D'^2  =  2x  -90 + 18x   = 0

 

20x - 90  = 0

 

20x  =90

 

x= 9/2

 

y = 15 - 3(9/2)  = 30/2 - 27/2  =  3/2

 

So x^2  + y^2   is a minimum  when  x= 9/2  and y =3/2

 

The minimum  is   (9/2)^2  + (3/2)^2   =   [ 81 + 9 ] / 4  =   90 / 4 =  45/2  =  22.5

 

cool cool cool

 
 Nov 22, 2020

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