The Gamma$(\alpha,\beta)$ distribution has pdf $f(x) = x^{\alpha-1}e^{-\beta x}$, with $E(X)=\frac{\alpha}{\beta}, Var(X) = \frac{\alpha}{\beta^2}$.
Find gamma distribution of maximal variance subject to $\alpha = \beta - 1$
My Answer:
Since $\alpha = \beta - 1$, we have $1-\beta + \alpha =0$
Then we have, $L=\frac{\alpha}{\beta^2}+\lambda(1-\beta+\alpha)$
We solve this by finding first order partial derivatives:
$\frac{\partial L}{\partial \alpha}=\frac{1}{\beta^2}+\lambda$
$\frac{\partial L}{\partial \beta}=\frac{-2\alpha}{\beta^3}-\lambda$
$\frac{\partial L}{\partial \lambda}=1-\beta+\alpha$
Then we set each equal to 0 and solve, giving $\alpha^*=1, \beta^*=2, \lambda^*=-\frac{1}{4}$
I felt that $-\frac{1}{4}$ was very small for maximal variance so wanted to check that I had not made a mistake.
You've overly complicated things, but the answer is not too wrong:
When $\alpha=\beta-1$, then $\mathsf{Var}(X)=\dfrac{\beta-1}{\beta^2}$ so we investigate where the derivative of the variance (w.r.t. $\beta$) is zero. $$0~{=\dfrac{\mathrm d (\beta-1)/\beta^2}{\mathrm d\beta}\\=\dfrac{(2-\beta)}{\beta^3}}$$
Which gives a solution for $\beta=2$ and $\alpha=1$. So the variance at this critical point is indeed $^+\tfrac 14$.
You can check that the critical point is a local maxima by peeking at its neighbourhood. $$\dfrac{0.9}{1.9^2}\lt\dfrac{1}{2^2}\gt \dfrac{1.1}{2.1^2}$$
When $\alpha=\beta-1$, then the variance is at most $1/4$.