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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

Best Answer 

 #2
avatar
+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:

 

 
 Aug 1, 2022
 #1
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0

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
avatar
+2
Best Answer

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:

 

 
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