# max value , two variables

#### idontknow

$$\displaystyle 1\leq t_1 \leq 4$$ and $$\displaystyle 2 \leq t_2 \leq 6 \;$$ ; $$\displaystyle \; t_1 , t_2 \in \mathbb{N}.$$
Find maximum value of $$\displaystyle M=\dfrac{\sin(t_1 t_2 ) }{t_1 t_2 }$$.

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#### mathman

Forum Staff
It is really a question of the max of $\frac{sin(x)}{x}$ where x ranges over the given values of $t_1t_2$.

idontknow

#### idontknow

It is really a question of the max of $\frac{sin(x)}{x}$ where x ranges over the given values of $t_1t_2$.
still cannot find the maximum .

#### idontknow

Now I got a result , is it correct ?
set $$\displaystyle t_1 t_2 =n$$ and $$\displaystyle A_n = \dfrac{ \sin(n) }{n} \; , \: 2\leq n=t_1 t_2 \leq 24$$.

$$\displaystyle \dfrac{A_{n+1}}{A_n} -1= \dfrac{n \sin(n+1) }{(1+n)\sin(n) }-1=\dfrac{n \sin(n+1)-\sin(n)-n\sin(n) }{(1+n)\sin(n) }<0\; , n\leq 24$$.

since $$\displaystyle A_n$$ is decreasing in the given domain then $$\displaystyle \sup A_n = A_2 = \dfrac{\sin 2}{2}$$.

#### mathman

Forum Staff
How do you know $nsin(n+1)-(n+1)sin(n) \lt 0$? $sin(x)$ oscillates!

#### romsek

Math Team
$$\displaystyle \sup A_n = A_2 = \dfrac{\sin 2}{2}$$.
As far as I can see from the graph the maximum occurs as close to $u = t_1 t_2 = 0$ as their domains allow, i.e.
$t_1 =1,~t_2 = 2 \Rightarrow u=2$

Which is what you have discovered in your own way.

idontknow