minimize $4x_1^2+2x_2^2-4x_1x_2-8x_2$
subject to $x_1+x_2\leq 4$
Does the point $(2,2)$ satisfy the FONC for a local minimizer?
The gradient of the objective function is
$\nabla f = \begin{bmatrix} 8x_1^2-4x_2-4x_2 \\ 4x_2^2-4x_1-8\end{bmatrix}$ = $\begin{bmatrix} 0\\0 \end{bmatrix} $
Filling in the point should give $0$ but it doesn't so the answer is no. Is this correct or do I have to use Lagrange?
Best Answer
Here is a non-calculus approach to find the minimum.
By completing the square we get $$ \begin{align} f(x_1,x_2)&=4x_1^2+2x_2^2-4x_1x_2-8x_2\\ &=4\left(x_1-\frac{x_2}2\right)^2+x_2^2-8x_2\\ &=4\left(x_1-\frac{x_2}2\right)^2+(x_2-4)^2-16\\ &=(2x_1-x_2)^2+(x_2-4)^2-16 \end{align} $$
After the substitution $y_1=2x_1-x_2$ and $y_2=x_2-4$ we transform this to the problem to minimize $$g(y_1,y_2)=y_1^2+y_2^2-16$$ for $$y_1+3y_2\le -4.$$
This is the same question as finding the point on the line $y_1+3y_2=-4$, which is closest to the origin. Geometrically we can see that this point must belong to the perpendicular line $3y_1-y_2=0$ and we get $y_1=-\frac25$, $y_2=-\frac65$ with $g(y_1,y_2)=\frac4{25}+\frac{36}{25}-16=\frac85-16=-\frac{72}5$.
This corresponds to $x_1=\frac65$ and $x_2=\frac{14}5$.
WolframAlpha yields the same result.