On how to find the maximum point of a formula. I need help finding out how to find the maximum of a formula I devised, given the starting conditions.
$$y = \sqrt{{\frac{{mx^2v^2 - 2xF}}{{m}}}}$$
or, adding in the values of F,
$$y = \sqrt\frac{{mx^2v^2 - xpcA(w - xv)^2}}{{m}}$$
No variable can equal 0, except for y. x should be undefined at 0.
When placing this is Desmos, it graphs a function with a maximum, I have found that,  it hits 0 at 2F/mv2. But I have yet to be able to find out when it hits the maximum and begins to decrease.
To further make myself clear, x is changing and I need to find a formula which says at what x, given the other variables, would y reach the maximum. Formula 2 is more relevant as it separates F into the different variables.
How exactly would I go about finding this maximum? I know how to do it with simpler functions, but how can I find a formula for the maximum of a more complicated one like this. I need to be able to find both the x and y of the maximum.
I am sorry if this is a lot of work, I am just not sure how to do things like this myself. At least I want to be taught how to be able to find the Maxima of complex functions, so that I can solve the problem myself.
Edit: Cleared up some confusion with variables, and specified my question better.
The Derivative I got
$$y = \frac{(cAxpw^2 - 2cAx^2 vpw + cAx^3v^2p)}{m^2}$$
setting to 0 gets $x = \frac{w}{v}$ and $x = 0$, which only seem to describe the zeros of the numerator of the derivative, and when placed into the original formula, does not get the correct value of x.
Edit: I am just giving up on this, no matter what I try I can't find the right answer, and it is proving stressful and taxing on my mental health. Need to stop focusing on it for my own sake.
 A: Let's write the formula as $$y=\sqrt{Ax^3+Bx^2+Cx}$$ where my $A,B,C$ are functions of your constants $m,v,p,c,A,w$. The cubic is zero at $x=0$ and also at the zeros of the quadratic $Ax^2+Bx+C$. Assuming the formula makes physical sense, and the cubic is positive for sufficiently small, positive values of $x$, the quadratic has two real zeros, one positive and one negative, and $y$ is positive between $x=0$ and the positive root of the quadratic, rising to a unique maximum in that interval (note that my $A$ is negative).
Now $y$ is maximized where $y^2$ is, so it's just a matter of differentiating the cubic, setting the resulting quadratic to zero, and solving for the positive value of $x$.
A: Function is maximum when its second derivative is less than zero. You need to check that condition first. After checking, you need to equate first derivative to zero and find x. Substitute this value of x in y and you will find answer.
A: I hope if some one view this in physical aspect then the first equation will become this( all the thing varying with respect to time) so why dt is used and after solving this we can get an equation in x and y w.r.t.t like f(x(t),y(t)) and it is hoped then we can find max or min as we do in multivariate calculus
y${\displaystyle\frac{dy}{dt}}$=x(${\displaystyle\frac{dx}{dt}}$)${^3}$+${x^2}$${\displaystyle\frac{dx}{dt}\frac{d^2 x}{dt^2}}$ - x${\displaystyle\frac{d^3 x}{dt^3}}$ - ${\displaystyle\frac{dx}{dt}\frac{d^2 x}{dt^2}}$
