x^2+2xy+2y^2+6y
= x^2+2xy+y^2 + y^2+6y
= (x+y)^2 + y(y+6) = S
x _ y _ S
0 0 0
1 0 1
2 0 4
3 0 9
4 0 16
...
0 1 8
0 2 20
.. -------no future here, S gets bigger
try negative values
0 -6 36
1 -6 25
..
5 -6 1
5 -5 -5
1 -1 -5
2 -2 -8
3 -3 -9 <-------- LOOKS LIKE THAT IS IT
4 -4 -8
Let S = x^2 + 2xy + 2y^2 + 6y
dS/dx = 2x + 2x y' + 2y + 4y y' + 6 y'
= 0 for a min of S
y' (2x + 4y + 6) = -2x - 2y
y' = (-2x-2y)/(2x+4y+6)
could it be when y = -x ?
If x and y are real numbers find the minimum value of x^2+2xy+2y^2+6y
1 answer