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.
+128707 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
