+140 Find the minimum value of
x^2 + xy + y^2 + x + y + 1
over all real numbers x and y.
Hi sherwyo!
Let \(F(x,y)=x^2+xy+y^2+x+y+1\)
Then,
\(F_x=2x+y+1\)
\(F_y=x+2y+1\)
Where \(F_x \text{ and, } F_y \) are the partial derivatives.
Now, since we want a minimum point, we set these partial derivatives to zero:
\(2x+y+1=0\) (1)
\(x+2y+1=0\) (2)
Multiply equation (2) by -2 and then add to equation (1), to get:
\(-3y-1=0 \implies y=-\dfrac{1}{3}\)
Thus,
\(x=-2y-1=\dfrac{2}{3}-1=-\dfrac{1}{3}\)
So the minimum is: \(F(-\frac{1}{3},-\frac{1}{3})=\dfrac{2}{3}\)
WolframAlpha:
Taking the derivatives with respect to x, y, we get
2x + y + 3 = 0
x + 2y + 3 = 0
Solving, we get x = y = -1. Therefore, the minimum is (-1)^2 + 1 + (-1)^2 + (-1) + (-1) + 1 = 2
Hi sherwyo!
Let \(F(x,y)=x^2+xy+y^2+x+y+1\)
Then,
\(F_x=2x+y+1\)
\(F_y=x+2y+1\)
Where \(F_x \text{ and, } F_y \) are the partial derivatives.
Now, since we want a minimum point, we set these partial derivatives to zero:
\(2x+y+1=0\) (1)
\(x+2y+1=0\) (2)
Multiply equation (2) by -2 and then add to equation (1), to get:
\(-3y-1=0 \implies y=-\dfrac{1}{3}\)
Thus,
\(x=-2y-1=\dfrac{2}{3}-1=-\dfrac{1}{3}\)
So the minimum is: \(F(-\frac{1}{3},-\frac{1}{3})=\dfrac{2}{3}\)
WolframAlpha:
