Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm working on a project that needs to solve a constraint optimization function. Currently, I'm using Knitro solver and it needs to calculate the the hessian of the lagrangian at x and lambda. I don't understand how to calculate the hessian. The constraint optimization problem is as follows: $\text{minimize}_x 100 -(x_2 - x^2_1)^2 + (1-x_1)^2$ subject to $1\le x_1x_2$, $ 0\le x_1,x_2$,$x_1\le0.5$. The function to calculate the hessian is (it seems that lambda is given): $ t = x_2 - x_1x_1;$ $ h_1 = (-400.0 t) + (800.0 x_1x_1) + 2.0$, $ h_2 = (-400.0 x_1) + \lambda_1$, $ h_3 = 200.0 + \lambda_2 2.0)$ Could you please tell me how to get $h_1,h_2,h_3$ ? Thank you very much.

share|improve this question
add comment

1 Answer

You don't need Lagrange multipliers to deal with this problem.

I read your constraints as $$0\leq x_1\leq {1\over2},\ x_2\geq0,\ x_1 x_2\geq1.$$ It follows that the feasible domain $B$ is bounded by the halfline $h_1: \ x_1={1\over2}$ starting upwards at the point $P:=({1\over2},2)$ and by a steep arc $h_2$ of the hyperbola $x_1 x_2=1$ starting at $P$ as well. The gradient of $f$ computes to $$\nabla f(x_1,x_2)=(\ldots, -2x_2+2x_1^2)\ .$$ It follows that $f$ has no stationary point in the interior of $B$, as $x_1\leq{1\over2}$ and $x_2\geq2$ there.

Along $h_1$ we have to consider the pullback $$\phi_1(x_2):=f\bigl({1\over2},x_2\bigr)=-x_2^2 +{1\over2} x_2+101.0625 $$ which obviously is monotonically decreasing to $-\infty$ going up along $h_1$. Analogously we can study the pullback along $h_2$, which is more complicated: $$\phi(x_2)=f\bigl({1\over x_2},x_2\bigr)=101-x_2^2 -{1\over x_2^4}+{1\over x_2^2}\ .$$ It should be possible to show that $\phi_2$ decreases monotonically to $-\infty$ as well when going upwards along $h_2$.

Since in fact we have $\lim_{x_2\to\infty}f(x_1,x_2)=-\infty$ uniformly in $x_1$ for $0\leq x_1\leq{1\over2}$ it follows that your function assumes its maximum at the point $P$ and is unbounded from below on the feasible domain $B$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.