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# Find the minimum value of a function with only real numbers as x and y

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Find the minimum value of

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

over all real numbers x and y.

Aug 1, 2022

#2
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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:

Aug 1, 2022

#1
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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

Aug 1, 2022
#2
+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:

Guest Aug 1, 2022