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Let x, y, and a be real numbers such that x+y=(a/2) and xy= -(a+3). Find the minimum possible value of x^2+y^2.

 Sep 9, 2023
 #1
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x + y = (a/2)     square  both sides

 

x^2 + 2xy + y^2 =  a^2/4

 

And

 

xy = -(a + 3)

2xy  = -2(a + 3)

 

So

 

x^2  - 2(a + 3) + y^2 = a^2/4

 

x^2 + y^2 =  a^2/4  - 2(a + 3)  =   (1/4)a^2 - 2a - 6        (1)

 

Take the derivative  of  (1)  and set    =  0 

 

(1/2)a  - 2  =  0

 

(1/2)a =  2

 

a =  4

 

The min value  of x^2 + y^2  =  (1/4)(4)^2  - 2(4)  - 6 =    4 - 8 - 6 =   -10

 

 

cool cool cool

 Sep 9, 2023
edited by CPhill  Sep 9, 2023
 #2
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It was 2. I figured it out. Take the derivitive of (1/4)a^2+2a+6. This is a/2+2=0. a=-4. Now plug it in.

Guest Sep 9, 2023

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