# Minimize $f(x,y)= e^{-xy}+\alpha e^{-1/x}$ with respect to $x$ for a given $y$

I am trying to minimize $$f(x,y)= \exp(-xy)+\alpha \exp\left(-\frac{1}{x}\right)$$ with reespect to $$x\in[0,\gamma]$$ for a given $$y\geq 0$$. Here $$\alpha$$ and $$\gamma$$ are positive constants.

Differnetiating $$f(x,y)$$ and solving does not provide any closed-form solution. I need to find a simple solution which is in some way related to the optimal value.

So, I tried replacing $$e^{-1/x}$$ with $$\ln(x)$$ as they both have similar behavior for certian range of $$x$$. I scaled $$\ln(x)$$ in a way that they both have same value at $$\gamma$$.

$$$$g(x,y)= \exp(-xy)+\alpha e^{-\frac{1}{\gamma}} \ln(x) / \ln(\gamma)$$$$

g(x,y) has a closed-form solution in terms of LambertW function. For cerain range of $$x$$, using plots, I see it to be very close to the optimal value of $$f(x.y)$$.

Now, I am trying to mathmatically show that they are indeed close.

Is there any good way to show this?

I tried the following

1) Derivative of $$f(x.y)$$ is close to zero at $$x^{*}$$ that minimizes $$g(x.y)$$

2) Show that both $$f(x.y)$$ an $$g(x.y)$$ are close for that range of $$x^*$$

• is there anything known about $\alpha$ or $\gamma$? I guess the result depends on their values as well – Jane Apr 29 at 4:50
• They both are positive constants. There are no other restrictions on them in general. I am trying to find the range of $\alpha$ for which this closeness works. – sar1729 Apr 29 at 5:02