max value , two variables

Dec 2015
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122
Earth
\(\displaystyle 1\leq t_1 \leq 4 \) 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
May 2007
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It is really a question of the max of $\frac{sin(x)}{x}$ where x ranges over the given values of $t_1t_2$.
 
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Dec 2015
935
122
Earth
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
May 2007
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759
How do you know $nsin(n+1)-(n+1)sin(n) \lt 0$? $sin(x)$ oscillates!
 

romsek

Math Team
Sep 2015
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USA
\(\displaystyle \sup A_n = A_2 = \dfrac{\sin 2}{2}\).
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.

Clipboard01.jpg
 
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