If A and B are acute angles such that 2A+2B=π , then what is the...

Given $2A + 2B = \pi$, which simplifies to $A + B = \dfrac{\pi}{2}$.

Since $A$ and $B$ are acute angles, $B = \dfrac{\pi}{2} - A$.

The expression to maximize is $\sin A \cdot \sin B$.

Substituting $B = \dfrac{\pi}{2} - A$, we get $\sin A \cdot \sin\left(\dfrac{\pi}{2} - A\right) = \sin A \cdot \cos A$.

Multiplying and dividing by $2$, we obtain $\dfrac{2\sin A \cos A}{2} = \dfrac{\sin 2A}{2}$.

The maximum value of $\sin 2A$ is $1$, which occurs when $2A = \dfrac{\pi}{2}$, or $A = \dfrac{\pi}{4}$.

Therefore, the maximum value of $\sin A \cdot \sin B$ is $\dfrac{1}{2} \times 1 = \dfrac{1}{2}$.

Answer: $\dfrac{1}{2}$