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# Minimum value

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Find the minimum value of

x^2 + 2xy + y^2 + x + y + 1

over all real numbers x and y.

Aug 3, 2022

#2
+1

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!

Aug 4, 2022

#2
+1

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!

Guest Aug 4, 2022