Find the maximum likelihood estimator for $\theta$ and $\delta$ considering a random sample of size $n$ from a population with distribution $$f(x)=\frac{1}{\theta}e^\frac{-(x-\delta)}{\theta},\quad x>\delta$$

I know $L(\theta,\delta)=\prod\_{i=1}^n \frac 1 \theta e^\frac{-(x-\delta)} \theta$. When I simplify, I get $$L(\theta,\delta)=\frac 1 {\theta^n}e^{-\frac 1 \theta \sum\_{i=1}^n x\_i}e^{\frac{n\delta} \theta}$$ Taking the natural log of this function, I get $$\ell(\theta,\delta)=-n\ln\theta - \frac 1 \theta \sum\_{i=1}^n x\_i+\frac{n\delta} \theta$$

Now, taking the partial derivatives for $\ell$ with respect to $\theta$ and $\delta$, I have the following: 

$$\frac{d\ell}{d\theta}=-\frac n \theta +\frac 1 {\theta^2}\sum\_{i=1}^n x\_i-\frac{n\delta}{\theta^2}$$

$$\frac{d\ell}{d\delta}=\frac n \theta$$
Setting these both equal to zero and with some manipulation, I found $\hat\delta=\bar x$ and $\hat\theta=0$, but this does not make sense, because an estimator cannot be $0$. I have seen other examples where similar issues arise, but I'm not sure how I should look at this in order to actually figure out the value of $\theta$ that maximizes the function.