Find the maximum value of $(12\sin x-9\sin^{2} x)$ 
The maximum value of $(12\sin x-9\sin^{2} x)$
  is equal to 
$a.)\ 3 \\
\color{green}{b.)\ 4}  \\
c.)\ 5 \\
d.)\ \text{none of these}$

As 
$-1\leq \sin x\leq 1 ,\\
12\sin x-9\sin^{2} x \\
=12-9=3 \\
$
But the answer given is $4.$
I am looking for a short and simple way.
I have studied maths up to $12$th grade.
Note : I can't use calculus.
 A: Hint: If you complete the square, you will get: 
$$12\sin x - 9\sin^2 x$$ 
$$= 4-4 + 12\sin x - 9\sin^2 x $$
$$= 4-2^2 + 2 \cdot 2 \cdot 3\sin x - (3\sin x)^2 $$
$$= 4-(3\sin x - 2)^2.$$
Now, it is easy to determine the maximum value of this expression.
A: The hint using completing the square is a great way of doing this problem. If you are less clever (like me), you might not think of this. In general, it is standard to find a maximum of a single-variable real-valued continuous function by setting the derivative of the function equal to zero. 
If $f(x)$ is the thing you want to maximize, then
$$f'(x) = 12\cos x - 18\sin x \cos x  =  0$$
$$(\cos x)(2 - 3 \sin x)=0$$
Our candidates for values of $x$ which maximize your function are therefore the solutions to either of the following equations
$$\cos x =0$$
$$ 2 - 3\sin x = 0$$
The first equation corresponds to values of $x$ that give $\sin x = 1$. 
The second equation corresponds to values of $x$ that give $\sin x = 2/3$
If you plug in $\sin x = 1$ and $\sin x = 2/3$ into your original function, you'll find that the former yields $3$ and the latter $4$. 
Therefore $4$ is the maximum. 
A: As you noted $\sin(x)$ takes on all values from $-1 \dots 1$.
So consider $f(y) = 12y -  9y^2$, where $y$ is in the range $-1 \dots 1$.
What is the maximum of $f$?  The $y^2$ coefficient is negative, so the maximum occurs at the vertex of the parabola.
The vertex of a parabola $f(x) = ax^2 + bx + c$ occurs at $x = \frac{-b}{2a}$, so the vertex of $12y - 9y^2$ is at $y=\frac{-12}{2 \times -9} = \frac{2}{3}$.
Note that $y=\frac{2}{3}$ is in the range of $-1 \dots 1$, so there is some $x$ such that $y = \sin(x) = \frac{2}{3}$.
So the maximum is $f\left(\frac 23\right) = 12\times\frac 23 - 9\times\left(\frac 23\right)^2 = 8 - 4 = 4$.
