How to prove that a point is a maximum point I have the following function:

$$
\Pi =\int_{0}^{z}[x_{1} + \alpha y + \alpha \frac{N-2}{2}y - \beta(z) - \gamma ( \beta(z) - \beta(y))](N-1)y^{N-2} dy $$

The first derivative with respect to $z$ is:

$$
\frac{\partial \Pi}{\partial z}=[ x_{1} + \alpha z + \alpha \frac{N-2}{2}z - \beta(z)] (N-1) z^{N-2} - (1 + \gamma) \beta'(z)z^{N-1}
$$

at $z = x_{1}$ it must be zero so that:

$$
[ x_{1} + \alpha x_{1} + \alpha \frac{N-2}{2} x_{1} - \beta(x_{1})] (N-1)  - (1 + \gamma) \beta'(x_{1})x_{1} = 0
$$

The solution is:

$$
\beta(z) = \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}
$$

What I want to prove is that the function $\Pi$ have a global maximum in $x_{1}$. Until now I think I've only obtained a necessary condition. Can you suggest me a way to prove it?

EDIT: $\beta(\cdot)$ is an increasing function.
The first order condition results in a differential equation in $\beta(x_1)$, 

$$\left[ x_{1} + \alpha x_{1} + \alpha \frac{N-2}{2} x_{1} - \beta(x_{1})\right] (N-1)  - (1 + \gamma) \beta'(x_{1})x_{1} = 0$$

$$\implies (1 + \gamma) x_{1}\beta'(x_{1}) = -(N-1)\beta(x_{1}) + (N-1)\left(1+a \frac{N}{2}\right)x_1$$

or generically $$g(x_1) y' = f_1(x_1)y + f_0(x_1)$$

Where $y \equiv \beta(x_1)$ and the other obvious mappings.

This has general solution 

$$y= \beta(x_1) = C\cdot  \exp \{F\} + \exp\{F\}\int \exp\{-F\} \frac{f_0(x_1)}{g(x_1)} dx_1, \;\;\; F\equiv \int \frac{f_1(x_1)}{g(x_1)}dx_1 $$

and doing the work results in 

$$ \beta(x_1) = \beta(z^*) = Cx_1^{-(N-1)/(1+\gamma)} + \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}x_1$$

I guess there exists a condition in the model that leads to $C=0$ so that we get the linear expression for the $\beta$ function

$$ \beta(z^*) = \frac{(N \alpha + 2)(N-1)}{2(N + \gamma)}x_1$$

Regarding whether $\Pi(z)$ has a global maximum at the stationary point $z^* = x_1$, working the 2nd derivative with respect to $z$, and evaluating it at $z^*=x_1$ is what one can do.  

Then, if I haven't done any trivial mistake, it appears that a necessary condiiton for the second derivative to be negative at $z^* = x_1$ is that $\alpha(1+\gamma) &lt;1$ and, given this, the condition for a global maximum at $z^*=x_1$is 

$$N  &gt; \frac{2+\gamma}{1-\alpha(1+\gamma)}$$

which appears to put a lower bound in the number of participants in the auction, which I guess is what $N$ measures. 

But this is the 2nd time I had to guess (and a third guess is that parameters are positive), validating @denesp comment that even though the OP issue is mathematical, providing context and information on what the various symbols stand for, is needed if one is to have a good-quality question to work with. After all, this context is what would make this an economics question and not a math question.