I have to maximize the following function -
Max A $C_1^{-m}/{-m}$ + (1-A) $C_2^{-m}/{-m}$
Subject to,
$C_1$ ≤ 5(1-x) + x
$C_2$ ≤ 3(1-x) + 7x
1≤x≤10
I wrote it as: L(x) = f(x) - $λ_1$($C_1$ - 5(1-x) + x) - $λ_2$($C_2$ - 3(1-x) + 7x) - $λ_3$(x-1) - $λ_4$(x-10)
Can I write last constraint as partioned into $λ_3$ and $λ_4$. Is there some other way to introduce such box constraints into the same problem?
On
Considering
$$ L(x,\lambda,\epsilon) = \lambda_1(C_1-5(1-x)-x-\epsilon_1^2)+\lambda_2(C_2-3(1-x)-7x-\epsilon_2^2)+\lambda_3(1-x-\epsilon_3^2)+\lambda_4(x-10-\epsilon_4^2) $$
with $\epsilon_k$ convenient slack variables, the stationary points are the solutions for
$$ \nabla L = 0 = \cases{4\lambda_1-4\lambda_2-\lambda_3+\lambda_4 = 0\\ C_1-5(1-x)-x-\epsilon_1^2 = 0\\ C_2-3(1-x)-7x-\epsilon_2^2 = 0\\ 1-x-\epsilon_3^2 = 0\\ x-10-\epsilon_4^2 = 0\\ \lambda_1\epsilon_1 = 0\\ \lambda_2\epsilon_2 = 0\\ \lambda_3\epsilon_3 = 0\\ \lambda_4\epsilon_4 = 0 } $$
and the feasible solutions are those observing $\epsilon_k^2 \ge 0$ for $k = 1,2,3,4$
$$ \left( \begin{array}{ccccccccc} x & \lambda_1 & \lambda_2 & \lambda_3 & \lambda_4 & \epsilon_1^2 & \epsilon_2^2 & \epsilon_3^2 & \epsilon_4^2\\ \frac{5-C_1}{4}& 0 & 0 & 0 & 0 & 0 & C_1+C_2-8 & \frac{C_1-1}{4} & -\frac{1}{4} (C_2+35) \\ \frac{C_2-3}{4} & 0 & 0 & 0 & 0 & C_1+C_2-8 & 0 & \frac{7-C_2}{4} & \frac{C_2-43}{4} \\ x & 0 & 0 & 0 & 0 & C_1+4 x-5 & C_2-4 x-3 & 1-x & x-10 \\ \end{array} \right) $$
Is your function constant? If I read correctly you have \begin{align*} \underset{\text{w.r.t}\; x\in \mathbb{R}}{\text{maximise}}\; &\;-A\,\dfrac{C_1^{-m}}{m} -(1-A)\dfrac{C_2^{-m}}{m} = \gamma\\ \text{subj.to}&\quad h_1(x)\leq 0\\ &\quad h_2(x)(x)\leq 0\\ &\quad \alpha\leq h_3(x) \leq \beta \end{align*} Maximizing a constant function under some constraints is called a satisfiability problem; any reachable value will attain the maximum (and minimum) of $\gamma$. Your method of putting $$\alpha\leq h_3(x) \leq \beta \;\iff\; \alpha - h_3(x) \leq 0\; \text{and} \; h_3(x) - \beta \leq 0 $$ Before putting the problem into Lagrangian form is the go-to solution in most cases if $f$ is non constant!