Is there any way to maximise the given function While working through a question , I stumbled across the following equation
$$ c = \frac {(2-a_0)+ \sqrt{-a_0(3a_0-4)}}{2} $$
We are required to maximise the function $c$ and find its corresponding value 
I plotted its graph on Desmos and found its maxima to be 1/3 however I have no clue how to arrive at this result
Also we are not allowed to use calculus (at least it is discouraged )
 How are we supposed to proceed with this?
 A: This answer makes a few assumptions on the form of the function, but requires no calculus for its execution. Consider the function $$f(x)=\frac{2-x+\sqrt{4x-3x^2}}{2}$$. It's domain is the interval $[0,4/3]$. Now suppose that you wanted to invert that function. Generally, if the function has a few maxima and minima, it is true that the function is not invertible, but one can write different branches that invert the function in a specific domain. If it is known that the function has one global maximum/minimum, the inversion procedure will only yield 2 branches (as it happens with inverting a parabola for example). Thus we can identify that point as the only point for which the inversion procedure is UNIQUE, or in more mathematical terms, the point for which the equation $f(x)=y$ has exactly one solution.
NOTE: (In principle) this procedure can be used to determine the global minimum/maximum of any function that possesses either or both.
Having established this prior knowledge/intuition, we proceed to find the value of x for which $f(x)=y$ has a unique solution.
We can rewrite this equation in terms of a quadratic polynomial in y:
$$x^2+(y-2)x+(y-1)^2=0 ~~~(1)$$
If we demand this equation have a unique solution, then the discriminant has to be zero, yielding:
$$\frac{y-2}{y-1}=\pm2\Rightarrow y=0~,~y=\frac{4}{3}$$
We reject the candidate $y=0$, since there is no real $x$ in the domain of $f$ that yields a root (this can be easily checked) and we conclude that the global maximum/minimum value of the function is $4/3$. We plug this value into $(1)$ and we obtain that $x=1/3$ is the position of the max/min. 
Finally to establish whether this is a global maximum or a minimum, we evaluate $f(1)=1<\frac{4}{3}$ and evidently the point we isolated is a global maximum of $f$.
A: Well, I would say that you use the right tool for the job! It's not obvious to me how to do it without calculus, which would be the following:


*

*You can discover for yourself that the domain of $c$ is $a_0\in[0,4/3].$ 

*Note that 
$$c'(a_0)=\frac{1}{4} \left(\frac{4-6 a_0}{\sqrt{-a_0 (3 a_0-4)}}-2\right). $$

*Setting this equal to zero and solving for $a_0$ yields $a_0=1/3.$

*Evaluating $c$ at the endpoints and the one critical point, we have
\begin{align*}
c(0)&=1 \\
c(1/3)&=4/3 \\
c(4/3)&=1/3.
\end{align*}
Therefore, the max occurs at $1/3$ and has a value of $4/3.$
