For $x, y \in \mathbb{R}$, the least value of the expression $x^2 + 2xy + 3y^2 - 6x + 6y$ is

Question

The problem statement : For $x, y \in \mathbb{R}$, the least value of the expression $x^2 + 2xy + 3y^2 - 6x + 6y$ is -

I want to solve this using differential calculus.

This is what I have tried so far :

I am trying to differentiate it in terms of $x$, but it turns out that I am not able to implicitly differentiate it.

It would be great if someone could explain to me the solution to this problem.

Answer 1

Take $F(x,y)=x^2+2xy+3y^2-6x+6y.$ Their derivatives are $$F_x=2x+2y-6,\,F_y=2x+6y+6,$$ which is zero only if $x=6$ and $y=-3$. Their second order derivatives are $$F_{xx}=2,\,F_{yy}=6,\,F_{xy}=F_{yx}=2,$$ so the Hessian is \begin{pmatrix} 2&&2\\2&&6\end{pmatrix} which is positive definite, so the point is a global minimum with value $-27$.

Answer 2

I solved the question without noticing that the user of the original post wanted a solution using calculus; just treat it as an alternative solution.

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Never forget the method of completing the square. $$ \begin{aligned} & x^2 + 2xy + 3y^2 - 6x + 6y \\ = {} & x^2 + (2yx - 6x) + 3y^2 + 6y \\ = {} & x^2 + 2x (y - 3) + (y - 3)^2 + 3y^2 + 6y - (y - 3)^2 \\ = {} & (x + y - 3)^2 + 2y^2 + 12y - 9 \\ = {} & (x + y - 3)^2 + 2(y^2 + 6y) - 9 \\ = {} & (x + y - 3)^2 + 2(y^2 + 6y + 9 - 9) - 9 \\ = {} & (x + y - 3)^2 + 2(y^2 + 6y + 9) - 27 \\ = {} & (x + y - 3)^2 + 2(y + 3)^2 - 27 \\ \geq {} & {-27}. \end{aligned} $$ Hence, on one hand, $x^2 + 2xy + 3y^2 - 6x + 6y \geq -27$ for real numbers $x$, $y$.

On the other hand, if $x$ and $y$ are taken to meet both the requirements that $$ x + y - 3 = 0 $$ and that $$ y + 3 = 0, $$ which imply that $x = 6$ and $y = -3$, then $$ x^2 + 2xy + 3y^2 - 6x + 6y \color{red}{=} -27. $$

Hence the minimum is $-27$.

Answer 3

I always propagate to use differential forms. A local extremum has a horizontal tangent space defined by the differential form

$$ 0 = \mathbb d \ (x^2 + 2 x y + 3 y^2 -6 x + 6 y) = 2 (x+y-3) \mathbb d x + (6 y + 2 x +6) \mathbb d y $$

    Solve[x + y - 3 == 0 && 3 y + x + 3 == 0,{x,y}]
    {x -> 6, y -> -3}
    (x^2 + 2 x y + 3 y^2 -6 x + 6 y)/.{x -> 6, y -> -3}
    -27