Find the minimum value of
x^2 + 2xy + y^2 + x + y + 1
over all real numbers x and y.
Hello Guest!
Let \(F(x,y)=x^2+2xy+y^2+x+y+1\)
Then,
\(F_x=2x+2y+1\)
\(F_y=2x+2y+1\)
Where \(F_x \text{ and, } F_y\) are the partial derivatives.
Now we set the partial derivatives to zero to find the minimum.
Thus,
\(2x+2y+1=0 \\ 2y+2y+1=0 \\ \text{But, these two equations are identical!} \\ \text{The idea is, we make one variable the subject of the equation} \\ \text{Then, we substitute into the original function F(x,y), and this will yield the minimum}\)
So: \(x=-y-\dfrac{1}{2} \\ x^2+2xy+y^2+x+y+1=(x+y)^2+(x+y)+1=(x+y)(x+y+1)+1\)
Notice, we simplified the original function F(x,y) by factorization so when we substitute:
\((-y-\frac{1}{2}+y)(-y-\frac{1}{2}+y+1)+1=-\frac{1}{2}(\frac{1}{2})+1=\frac{3}{4}\) which is the minimum.
I hope this helps! And If you need further explanation, don't hesitate to ask!
Hello Guest!
Let \(F(x,y)=x^2+2xy+y^2+x+y+1\)
Then,
\(F_x=2x+2y+1\)
\(F_y=2x+2y+1\)
Where \(F_x \text{ and, } F_y\) are the partial derivatives.
Now we set the partial derivatives to zero to find the minimum.
Thus,
\(2x+2y+1=0 \\ 2y+2y+1=0 \\ \text{But, these two equations are identical!} \\ \text{The idea is, we make one variable the subject of the equation} \\ \text{Then, we substitute into the original function F(x,y), and this will yield the minimum}\)
So: \(x=-y-\dfrac{1}{2} \\ x^2+2xy+y^2+x+y+1=(x+y)^2+(x+y)+1=(x+y)(x+y+1)+1\)
Notice, we simplified the original function F(x,y) by factorization so when we substitute:
\((-y-\frac{1}{2}+y)(-y-\frac{1}{2}+y+1)+1=-\frac{1}{2}(\frac{1}{2})+1=\frac{3}{4}\) which is the minimum.
I hope this helps! And If you need further explanation, don't hesitate to ask!
