Given min $1/2x^TQx+d^Tx$ s.t $a_1^Tx\leq b_1$
$a_2^Tx\leq b_2$
and $Q\succ0(n\geq3)$ $a_1,a_2,d\in\mathbb{R}^n$ and $b_1,b_2\in\mathbb{R}$,Assume that $a_1^TQ^{-1}a_1=a_2^TQ^{-1}a_2=2,a_2^TQ^{-1}a_1=0,a_1,a_2\neq0$
Find the optimal solution
I've set the lagrangian to be $L(x,\lambda,\eta)=1/2x^TQx+d^Tx+\lambda(a_1^Tx- b_1)+\eta(a_2^Tx-b_2)$
$\frac{\partial L}{\partial x}=Qx+d+\lambda a_1+\eta a_2$
I've tried setting $\lambda\neq0,\eta=0$ and solving but didn't get me anywhere,any hint please?
So you got $L(x,\lambda,\eta)=1/2x^TQx+d^Tx+\lambda(a_1^Tx- b_1)+\eta(a_2^Tx-b_2)$ and derived the stationarity condition: $$Qx+d+\lambda a_1+\eta a_2=0$$ You somehow need to use that $a_1^TQ^{-1}a_1=a_2^TQ^{-1}a_2=2,a_2^TQ^{-1}a_1=0,a_1,a_2\neq0$. If you pre-multiply the stationarity condition with $a_1^TQ^{-1}$ and $a_2^TQ^{-1}$ you get: $$a_1^Tx+a_1^TQ^{-1}d+2\lambda=0 \quad \text{and} \quad a_2^Tx+a_2^TQ^{-1}d+2\eta=0$$ which means: $$\lambda=-0.5(a_1^Tx+a_1^TQ^{-1}d) \quad \text{and} \quad \eta=-0.5(a_2^Tx+a_2^TQ^{-1}d)$$ Plugging that back into the stationarity condition, you get: $$Qx+d-0.5(a_1^Tx+a_1^TQ^{-1}d) a_1-0.5(a_2^Tx+a_2^TQ^{-1}d) a_2=0$$ Since $(a_1^Tx)a_1 = a_1(a_1^Tx)=(a_1a_1^T)x$, the stationarity condition can be simplified to: $$(Q-0.5a_1a_1^T-0.5a_2a_2^T)x+d-0.5a_1^TQ^{-1}d a_1-0.5a_2^TQ^{-1}d a_2=0$$ so $$x = (Q-0.5a_1a_1^T-0.5a_2a_2^T)^{-1}(0.5a_1^TQ^{-1}d a_1+0.5a_2^TQ^{-1}d a_2-d)$$