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Given that 5x + 12y = 13, find the minimum value of x^2 + y^2.

 May 2, 2020
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5x + 12y  = 13

12y  = 13 - 5x

y  =  [ 13 - 5x] / 12

 

x^2 + y^2   =

 

x^2  + [  (13 -5x ) /12  ]^2

 

x^2  + (1/144) [ 13 - 5x ]^2

 

Take the derivative of this and set to 0

 

2x  +  (2/144) [ 13 - 5x ] (-5)  =  0      simplify

 

2x - (10/144) [ 13 - 5x ]  = 0

 

2x - (5/72] [ 13 - 5x ]  =  0

 

2x   =   (5/72)[ 13 -5x ]

 

(144/5)x  =  13 - 5x

 

(144/5)x + 5x  = 13

 

[ (144 + 25 )/ 5 ] x  = 13

 

[169/5] x  = 13

 

x = [ 13 * 5 ] / 169

 

x =  5/13

 

y = [ 13 - 5 ( 5/13) ] / 12   =   [ 13 - 25/13] / 12   =  [ 169 - 25] / 156  =  144 / 156  =  12/13

 

So.....

 

min  [ x^2 + y^2 ]    =   

 

[5/13]^2  + [12/13]^2     =   

 

[ 5^2 + 12^2 ] / 13^2  = 

 

 [ 25 + 144 ] / 169  = 

 

169/169  = 

 

1

 

 

 

cool cool cool

 May 2, 2020

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