If $\alpha\,\, and\,\, p$ are non-negative scalars, are the two problems given below equivalent? I know that the values of objective functions will be different but I want to know will they result in same values of $\alpha\,\, and\,\, p$:
\begin{align} Problem 1: \min_{\alpha_{1},p_{1},\alpha_{2},p_{2}}\alpha_{1}p_{1}+\alpha_{2}p_{2},\\ \text{subject to}: \alpha_{1}log(1+p_{1} )\geq5,\\ \alpha_{2}log(1+p_{2})\geq5, \alpha_{1}\,\, and\,\, \alpha_{2}\leq1.\\ Problem 2: \min_{\alpha_{1},p_{1},\alpha_{2},p_{2}}\alpha_{1}^{2}p_{1}+\alpha_{2}^{2}p_{2},\\ \text{subject to}: \alpha_{1}log(1+p_{1} )\geq5,\\ \alpha_{2}log(1+p_{2})\geq5,\alpha_{1}\,\, and\,\, \alpha_{2}\leq1 .\\ \end{align}
No, they are not. After the obvious reduction the problems are
Problem 1: $$ \min 2ap\quad\text{subject to } a\ge\frac{5}{\ln(1+p)},\ p> 0 $$ Problem 2: $$ \min 2a^2p\quad\text{subject to } a\ge\frac{5}{\ln(1+p)},\ p> 0 $$ The minimum w.r.t. $a$ is clearly for equality in both cases $$ a=\frac{5}{\ln(1+p)} $$ then we have
Problem 1: $$ \min_{p>0} \frac{10p}{\ln(1+p)}. $$ Problem 2: $$ \min_{p>0} \frac{50p}{\ln^2(1+p)}. $$ The first problem has no solution ("optimal" when $p=0$, $a=+\infty$), the second problem has a solution.
The green plot: $\frac{x}{\ln(1+x)}$, the red plot: $\frac{x}{\ln^2(1+x)}$.