Find the critical points of f(x, y) :
[tex]\dfrac{\partial f}{\partial x}=2(x-1)-2(3-x-2y)=4x+4y-8=0[/tex]
[tex]\dfrac{\partial f}{\partial y}=2(y-4)-4(3-x-2y)=4x+10y-20=0[/tex]
Subtract the first equation from the second to eliminate x and solve for y :
[tex](4x+10y-20)-(4x+4y-8)=0\implies 6y=12\implies y=2[/tex]
Solve for x :
[tex]4x+4\cdot2-8=0\implies 4x=0\implies x=0[/tex]
So f(x, y) has one critical point at (0, 2).
Compute the Hessian determinant of f(x, y) at this point:
[tex]\mathbf H(x,y)=\begin{bmatrix}\frac{\partial^2f}{\partial x^2}&\frac{\partial^2f}{\partial x\partial y}\\\frac{\partial^2f}{\partial y\partial x}&\frac{\partial^2f}{\partial y^2}\end{bmatrix}=\begin{bmatrix}4&4\\4&10\end{bmatrix}[/tex]
The Hessian has determinant 24 > 0, which indicates a minimum, so the minimum value of f(x, y) is f(0, 2) = 6.
