The discussion focuses on finding the maximum and minimum values of the function $sin(x)sin(y)$ under the constraint $x+y=\dfrac{2\pi}{3}$, with the bounds $0 \leq x \leq \dfrac{\pi}{2}$. The maximum value occurs when $x = \dfrac{\pi}{3}$ and $y = \dfrac{\pi}{3}$, yielding a maximum of $\dfrac{3}{4}$. Conversely, the minimum value is achieved at the endpoints of the interval, specifically when $x = 0$ or $x = \dfrac{\pi}{2}$, resulting in a minimum of 0.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in optimization problems involving trigonometric functions.
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:(*)=
$\sin\,x \sin\, y = \sin ^2 \frac {\pi}{3} - \sin ^2 \alpha = \frac{3}{4} - \sin ^2 \alpha$-----(*)