# How can CVX solve the maximization problem "log(1-y/x)" with "x>0" and "y>0"

How can CVX solve the problem “max f(x,y) = log(1-y/x)” with “x>0” and “y>0”? Here, f(x,y) is joint concave about x and y. Thank you.

This is neither jointly concave nor convex in x and y for x >= 0, y >= 0, your claim to the contrary notwithstanding… For instance, its Hessian has one positive and one negative eigenvalue at x=2, y = 1.

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Really thanks for your kindly reply. If another constraint “x > y” is added to this problem, I think the objective function is concave. Then, could I use CVX to solve this problem?

Ii is neither convex nor concave for x > y > 0 or y > x > 0.

Please slow down a little and carefully read the link in my previous post.

Heuristic: \log(1-y/x)=\log(x-y) - \log(x) and if x,y are two independent variables then x-y,x are equally independent by a linear transformation, so your function is basically of the form \log(a)-\log(b) for two independent variables a,b. That is neither convex nor concave. It might be a different story in some cases if x,y are related in some way.

Thanks for your reply. I’m inspired by your ideas. In my considered system, x and y are indeed related, and the original problem can be transformed into a form as “log(1+(C-y)/x)”, where C is a constant parameter and C > 0, with a constraint “y < x + C”. Based on these conditions, its Hessian has two negative eigenvalues. Really thanks for your kind reminder.

Thanks again. I will double check and carefully read the link.