<s>[INST] 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.
 [/INST] 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 $f\left(\frac 23\right) = 12\times\frac 23 - 9\times\left(\frac 23\right)^2 = 8 - 4 = 4$.
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