min $x^2 -xy +y^2 -5x+6y$ subject to $1 \leq y$, $y^3 \leq 2x$, and $x \leq 8$
Write out the KKT conditions for this problem. Show that $(x,y) = (4,2)$ is a KKT point, and is therefore a global minimizer. Then show that $ y \to y^3 $ is convex for $y \geq 1$, show that this is a convex programming problem.
$L(x,y,\lambda_1,\lambda_2,\lambda_3)=x^2 -xy +y^2 -5x+6y-\lambda_1(y-1)-\lambda_2(y^3-2x)-\lambda_3(x-8)$
KKT Conditions: $$\nabla _xL(x,y,\lambda_1,\lambda_2,\lambda_3)=2x-y-5+2\lambda_2 - \lambda_3=0 $$
$$\nabla _yL(x,y,\lambda_1,\lambda_2,\lambda_3)= -x+2y+6-\lambda_1-3\lambda_2y^2=0$$
$$ \lambda_1,\lambda_2,\lambda_3> 0$$
$$1 \leq y, y^3 \leq 2x,x \leq 8$$
$$\lambda_1(y-1)=0$$ $$\lambda_2(y^3-2x)=0$$ $$\lambda_3(x-8)=0$$
Now to verify that $(4,2)$ is a KKT point: All I have to do is plug in (4,2) to $\nabla _xL$ and $\nabla _yL$ to find $ \lambda_1,\lambda_2,\lambda_3$ but I have three unknowns and only 2 equations so how do I find the $\lambda$'s.
Any suggestions of how to show that $ y \to y^3 $ is convex for $y \geq 1$, and that this is a convex programming problem... I'm not sure how to go about this part.