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\begin{equation} \begin{aligned} \min_{t_1, t_2 \in R} \quad & t_1t_2 + \frac{1}{t_1t_2^2}\\ \textrm{s.t.} \quad & t_1, t_2 > 0\\ \\ \end{aligned} \end{equation}

How do I make the objective value of the given primal program arbitrarily close to zero?
As $t_1$ and $t_2$ both have inverse ($\frac{1}{t_1}$ and $\frac{1}{t_2^2}$ respectively), can't make both terms go zero..

jun
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2 Answers2

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Fix $t_2=a$, and consider minimising $at_1 + \frac{1}{a^2 t_1}$ for $t_1>0$. Verify its minimum is $\frac{2}{\sqrt{a}}$ at $t_1=\frac{1}{a^{3/2}}$. So for the solution $(t_1,t_2)=(\frac{1}{a^{3/2}}, a)$, the value of the function is $\frac{2}{\sqrt{a}}$. Now let $a\rightarrow \infty$.

For example, $(t_1, t_2)=(\frac{1}{100000^{1.5}}, 100000)$, the function value is $\approx 0.006324...$

AspiringMat
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Consider the line $t_2 = \frac{1}{t_1^{2/3}}$

Now, if we let $t_1 \to 0$ along this line, then we can see that the value of the function is asymptotic to zero, since:

$$t_1t_2 + \frac{1}{t_1t_2^2} = t_1^{1/3} + t_1^{1/3} = 2t_1^{1/3}$$

Sean Lee
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