Consider a time independent projectile motion thrown from the ground from a height of zero:
$$ y = x \tan \theta - \frac{g}{2v_o^2} \sec^2 \theta x^2$$
Where $x$ is a horizontal range, $y$ is the height for a given $x$ and $ \theta$ is the angle which the projectile is thrown at it, $u_x$ and $u_Yy$ are horizontal and vertical velocities respectively.
Figuring out maximum height using implicit function theorem:
$ (x_{max} , 0)$ is a solution to the above curve, (ground is reference):
$$F= 0 = y_o + x_{max} \tan \theta - \frac{g}{2 v_o^2} \sec^2 \theta x^2$$
Now, apply the implicit function theorem:
$$ \frac{dx}{d \theta} =- \frac{\frac{\partial F}{\partial \theta}}{ \frac{\partial F}{\partial x}}$$
For optimum conditions,
$$ \frac{ \partial F}{\partial \theta} = 0$$
Or,
$$ 0=x_{max} \sec^2 \theta - \frac{g}{2 v_o^2} 2 \sec^2 \theta \tan \theta x_{max}^2$$
Simplifying
$$ 0 = 1 - \frac{g}{v_o^2} \tan \theta x_{max}^2$$
$$ \sqrt{\frac{ v_o^2}{g \tan \theta} }= x_{max}$$
Seems so that there is no real 'optimal' angle of throw, you can make theta as small as you want to make the projectile go further.
Another way:
The time of flight is given as:
$$ t = 2 \frac{u_y}{g}$$
And, the horizontal displacement equation:
$$ x_{max} = u_x t =\frac{ 2 u_y u_x}{g}$$
Now, my question is why does each way give a different answer? Are they both right or have I made a mistake somewhere?