My problem:
\begin{subequations}
\begin{align}
\mathop{\max} & \quad x\
\text { s.t. }
& 0\leq x\leq1\
& -\frac{2}{3}x^6+x^5<=5.
\end{align}
\end{subequations}
My code:
cvx_begin
variables x2
maximize x2
subject to
x2<=1;
x2>=0;
poly_env([-0.6667 1.0000 0 0 0 0 0],x2)<=5;
cvx_end
end
-\frac{2}{3}x^6+x^5 is convex when x\in \left[0,1\right], but the cvx tool think it concave. Is there any way to correct my problem.
-2/3*x^6+x^5 is convex for 0 <= x <= 1, but it is not convex for the entirety of its natural domain (-Inf <= x <= inf). I don’t know of a way of representing it in CVX. I am not ruling out that a clever forum reader might come up with a way, but I don’t think it is likely.
it’s not that CVX thinks it is concave. if -2/3*x^6+x^5 is entered directly, CVX concludes that it violates its rules, and it can’t determine its curvature (convex, concave, or neither). It also violates the rules for use of poly_env.
Presuming you actually want to maximize this function, it is a concave programming problem with compact constraint region. Therefore, there is a global maximum at an extreme, of which there are only two, namely the endpoints, 0 and 1. Therefore, whichever of 0 and 1 has the larger objective value is the global optimum. If you actually wished to minimize it, a local nonlinear solver such as FMINCON should find the global minimum, because the optimization problem would be convex (due to being constrained to [0,1]).
Thanks for your quick reply! It helps a lot.