The Lagrangian for this problem is
[tex]L(x,y,\lambda)=x^2+3y^2+\lambda(x+y-1)[/tex]
and has partial derivatives
[tex]\begin{cases}L_x=2x+\lambda\\L_y=6y+\lambda\\L_\lambda=x+y-1\end{cases}[/tex]
Set each partial derivative equal to 0 and solve for [tex]x[/tex] and [tex]y[/tex]:
[tex]\begin{cases}2x+\lambda=0\\6y+\lambda=0\\x+y=1\end{cases}[/tex]
Subtracting the second equation from the first, we get
[tex]2x-6y=0\implies x-3y=0[/tex]
and subtracting this from the third equation yields
[tex]4y=1\implies y=\dfrac14[/tex]
which means
[tex]x+\dfrac14=1\implies x=\dfrac34[/tex]
So a critical point occurs at [tex]\left(\dfrac34,\dfrac14\right)[/tex] (or (0.750, 0.250)). The minimum value would then be [tex]f\left(\dfrac34,\dfrac14\right)=\dfrac34=0.750[/tex].
