Maxima & Minima

Taking $\nabla f(x,y) = 0$ , we obtain:

$f_{x} = 4y + 6x +3x^2=0$ (i)

$f_{y} = 2y + 4x = 0$ (ii).

Knowing that $y = -2x$ from (ii), substituting this value into (i) produces the quadratic $3x^2 - 2x = 0 \Rightarrow x = 0, \frac{2}{3}$ . This gives us two critical points to consider: ( $x,y) = (0,0); (\frac{2}{3}, -\frac{4}{3}).$ The Hessian matrix of $f$ calculates to:

$F(x,y) = \begin{bmatrix} 6 + 6x & 4 \\ 4 & 2 \end{bmatrix}$ (iii)

and plugging in the two above critical points yields:

$F(0,0) = \begin{bmatrix} 6 & 4 \\ 4 & 2 \end{bmatrix} \Rightarrow$ indefinite matrix (aka saddle point),

$F(\frac{2}{3},-\frac{4}{3}) = \begin{bmatrix} 10 & 4 \\ 4 & 2 \end{bmatrix} \Rightarrow$ positive-definite matrix (aka global minimum point).

Hence, the minimum of $f$ computes to $f(\frac{2}{3},-\frac{4}{3}) = \boxed{-\frac{4}{27}}.$