If x and y are real numbers find the minimum value of x^2+2xy+2y^2+6y
look at my attempt here
http://www.jiskha.com/display.cgi?id=1455132924
To find the minimum value of the expression x^2 + 2xy + 2y^2 + 6y, we can use calculus.
1. Take the partial derivatives of the expression with respect to x and y:
∂(x^2 + 2xy + 2y^2 + 6y)/∂x = 2x + 2y
∂(x^2 + 2xy + 2y^2 + 6y)/∂y = 2x + 4y + 6
2. Set these partial derivatives equal to zero and solve for x and y:
2x + 2y = 0 --> x = -y
2x + 4y + 6 = 0
3. Substitute the value of x from the first equation into the second equation:
2(-y) + 4y + 6 = 0
2y + 6 = 0
y = -3
4. Substitute the value of y back into the first equation to find x:
x = -(-3) = 3
So, the minimum value of the expression x^2 + 2xy + 2y^2 + 6y occurs when x = 3 and y = -3.