# A nonlinear optimization problem

Can you please help me how can solve the following nonlinear program by indicating some useful reference or videos relevant to this problem? With my limited knowledge in this field, I tried to solve it using the Karush-Kuhn-Tucker (KKT) conditions, but it is not working.

\begin{alignat*}{2} \text{maximize } \qquad & f(x_1, x_2 , x_3 , x_4, x_5, x_6)=\Big[8\left(x_1^{3} + x_2^{3} + x_3^{3} + x_4^{3} + x_5^{3} + x_6^{3}\right) \Big]^{\frac{2}{3}} \\ \text{subject to } \qquad & 4\left(x_1^{2} + x_2^{2} + x_3^{2} + x_4^{2}\right) \le 1\\ & 4\left(x_3^{2} + x_4^{2} + x_5^{2} + x_6^{2}\right) \le 1\\ & 4\left(x_1^{2} + x_2^{2} + x_5^{2} + x_6^{2}\right) \le 1\\ & x_1, x_2, x_3, x_4, x_5, x_6 \ge 0. \end{alignat*}

• Show us your trial to see where it fails – paf Jun 16 '18 at 10:35
• Are you sure this is convex? WA suggests it is not. – Rodrigo de Azevedo Jun 16 '18 at 10:37
• Isn't it $f(x_1, x_2 , x_3 , x_4, x_5, x_6)=\Big[8\left(x_1^{2} + x_2^{2} + x_3^{2} + x_4^{2} + x_5^{2} + x_6^{2}\right) \Big]^{\frac{2}{3}}$ instead? – Rodrigo de Azevedo Jun 16 '18 at 10:43
• Prof. Rodrigo de Azevedo thank you very much for your answer, but I am interesting for the above conjecture object function is the aim, but I am not checking if it is convex or not? so building upon your answer, is this means that my nonlinear problem is not working due to the objective function is not convex? – Marcus Jun 16 '18 at 14:06
• Note that you are maximizing the objective, so you don't want it to be convex, you want it to be concave (which it isn't as it is convex quadratic when you set all but one variable to 0). Also note that the power $2/3$ is completely redundant as that is a monotonic increasing operation. – Johan Löfberg Jun 16 '18 at 14:25

$$f=\left(8\sum_{k=1}^6x_k^2\right)^{\frac{2}{3}}\leq(4\cdot\frac{3}{4})^{\frac{2}{3}}=\sqrt[3]{9}.$$ The equality occurs for $x_k=\frac{1}{4},$ which says that we got a maximal value.