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I have the function

$m(x, y)=x^2+y^2+xy+3y+13, R(ρ)=\{x, y\}\ \in \mathbb{R}^2: -ρ\le\ x, y \le\ ρ$

I wanna show that if ρ>2 so that m has a local minimum within R(ρ) then this minimum is absolute and will always be inferior to any local minimum that is on the border of R(ρ). It is the last part of an exercise, it does not ask for a rigorous proof just a justification, but i am really lost.

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  • $\begingroup$ So what you've tried? If it is an exercise in multivariable calculus, then you may first find the local minimum without constraint, and then you could use the Lagrange multipliers to find the minimum and maximum on the boundary. $\endgroup$
    – xbh
    Dec 28, 2017 at 2:03

2 Answers 2

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$\frac {\partial m}{\partial x} = 2x + y = 0\\ \frac {\partial m}{\partial y} = x + 2y + 3 = 0\\ (x,y) = (1,-2)$

Is a critical point.

$m(x,y) = (x,y) \pmatrix {1&\frac 12\\ \frac 12 & 1} \pmatrix {x\\y} + 3y + 13$

That matrix is positive definite (has two positive eigenvalues) which means that $x,y$ moving away from the critical point only makes $m$ get bigger.

Or,

$m(x,y) = \frac 34 ((x-1)+(y+2))^2 + \frac 14((x-1)-(y+2))^2 +10$

Which clearly has a minimum at $(1,-2)$

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  • $\begingroup$ How did u find this matrix? $\endgroup$
    – Desperados
    Dec 28, 2017 at 2:16
  • $\begingroup$ It is standard for quadratic forms $\pmatrix {(x^2) & \frac 12 (xy) & \frac 12 (xz)\\\frac 12(xy) & (y^2) & \frac 12 (yz)\\\frac 12(xz) & \frac 12(yz) & (z^2)}$ where $(x^2),(xy)\cdots$ are the coefficients of corresponding terms in the quadratic. $\endgroup$
    – Doug M
    Dec 28, 2017 at 2:21
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$ x^2+y^2+xy+3y+13 = \left(x+\frac{y}{2}\right)^2 + \frac{3}{4} (y + 2)^2 + 10 \ge 10 $ with equality iff $y=-2, x=1$.

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