The discussion revolves around finding the maximum and minimum values of the function $sin(x)sin(y)$ under the constraint $x+y=\dfrac {2\pi}{3}$, with the additional condition that $0\leq x\leq \dfrac{\pi}{2}$. The focus is on mathematical reasoning related to optimization within specified bounds.
The discussion does not present any disagreement or competing views, but it remains unresolved as no solutions or methods have been provided to address the problem.
There are no specific limitations or assumptions discussed, but the problem is framed within the context of the given constraints.
Individuals interested in mathematical optimization, particularly in the context of trigonometric functions and constraints.
Albert said:$x+y=\dfrac {2\pi}{3}$
$0\leq x\leq \dfrac{\pi}{2}$
$find: $
$(1):max(sin(x)sin(y))$
$(2):min(sin(x)sin(y))$
kaliprasad said:we have
$\sin (A+B) \sin (A-B) = (\sin\, A \cos\, B + \cos \, A \sin \, B) (\sin\, A \cos\, B - \cos \, A \sin \, B)$
$=(\sin^2 A \cos^2 B - \cos^2 A \sin^2 B) = \sin ^2 A (1- sin ^2 B ) - \sin ^2 B ( 1 - sin ^2 A) = \sin ^2 A - \sin ^2 B$
let $x= \frac {\pi}{3} + \alpha$ and hence $y = \frac {\pi}{3} - \alpha$
so
$\sin\, \sin\, y = \sin ^2 \frac {\pi}{3} - \sin ^2 \alpha = \frac{3}{4} \sin ^2 \alpha$-----(*)
the above is maximum when $\sin^2 \alpha = 0$ minimum or $x = y = \frac {\pi}{3}$ and value is $\frac{3}{4}$
now both A and B are 0 or positive and minumum is zero when $x= 0, y = \frac{2\pi}{3}$
Albert said:(*)=
$\sin\,x \sin\, y = \sin ^2 \frac {\pi}{3} - \sin ^2 \alpha = \frac{3}{4} - \sin ^2 \alpha$-----(*)