The beta distribution can have its parameter estimated via method of moments, which I will be doing.
$$\hat\alpha = \bigg(\dfrac{\bar x (1-\bar x)}{var(X)} - 1\bigg)\bar{x}\\ \hat\beta = \bigg(\dfrac{\bar x (1-\bar x)}{var(X)} - 1\bigg)(1-\bar{x}) $$
I have the mean but not the variance. How can I calculate the range of variance values that result in possible $(\hat\alpha, \hat\beta)$ vectors (both components positive)?
My attempt was with some calculus to find maxima and minima, and it did not go well.
$$ \hat\alpha(v\vert \bar x) = \bigg(\dfrac{\bar x (1-\bar x)}{v} - 1\bigg)\bar{x} = \dfrac{\bar x^2 (1-\bar x)}{v} - \bar x\\ \dfrac{d\hat\alpha}{dv} = -\dfrac{\bar x^2 (1-\bar x)}{v^2} = 0\\ \bar x^2(1-\bar x) = 0\\ \bar x = 0, 1 $$ My $\bar x$ is not going to be $0$ or $1$, and this approach loses the variance $v$.
(The reason I have the mean but not the variance is because I am using the beta as a prior for the $p$ parameter of a binomial variable, and I know what I want as the mean of my prior, but I do not know the variance. I want to know what variance values I can pick and if suggested values will be off-limits.)
