# Which of the following optimization problems will have a higher optimal value?

I have following optimization problems. The first problem is $$\underbrace{\max}_{x_1,x_2} \frac{\ln(1+a_1x_1)+\ln(1+a_2x_2)}{x_1+x_2+c}\\ \text{s.t. } 0\leq x_1\leq \bar{x}\\ 0\leq x_2\leq \bar{x}$$ and the second problem is $$\frac{\underbrace{\max}_{x_1,x_2}\ln(1+a_1x_1)+\ln(1+a_2x_2)}{x_1+x_2+c} \\ \\ \text{s.t.} \\0\leq x_1\leq \bar{x}\\ 0\leq x_2\leq \bar{x}$$ where $a_1,a_2$ and $c$ are some positive constants. $\bar{x}$ is some positive value which $x_1$ and $x_2$ cannot exceed. Which of the problem will have a higher optimized value? First or second? Any help in this regard will be much appreciated. Thanks in advance.

• What is $\bar{x}$? The mean of $x_1$ and $x_2$? – YukiJ Apr 10 '18 at 6:25
• @YukiJ no its a bound on the maximum value of $x_1$ and $x_2$. I have also added it in my question. – Frank Moses Apr 10 '18 at 6:27
• @OnceUponACrinoid you mean in the same expression for $x_1$ i use different values of $x_1$ in the numerator and different value for $x_1$ in the denominator? Is it possible? – Frank Moses Apr 10 '18 at 6:43
• What is the value of $x$ in the denominator? – LinAlg Apr 10 '18 at 16:03

I am presuming that $x_1$ and $x_2$ in the denominator of the 2nd problem are the respective argmaxes, $x_1^*$ and $x_2^*$,of the first problem, and therefore are input data, not optimization variables, for the 2nd problem.. Let $x_1^{**}$ and $x_2^{**}$ be the argmaxes of the 2nd problem. Then because the constraints are identical, obviously $$\frac{\ln(1+a_1x_1^*)+\ln(1+a_2x_2^*)}{x_1^*+x_2^*+c} \ge \frac{\ln(1+a_1x_1^{**})+\ln(1+a_2x_2^{**)}}{x_1^*+x_2^*+c}$$ becauee the 2nd problem has in effect been suboptimized.
• thank you for your answer. What if we put $x_1=x_1^{**}$ and $x_2=x_2^{**}$ in the second problem?(please comment) Actually we have to pick $x_1$ and $x_2$ in the second problem such that the numerator is maximized in the second problem. While in the first problem we have to pick $x_1,x_2$ such that the ratio is maximized. – Frank Moses Apr 10 '18 at 23:20