Answer
The maximum is $f(\pi/4) =0.5$ and the minimum is $f( 0 ) =f(\pi/2)=0$
Work Step by Step
Given
$$y=\sin x \cos x, \quad\left[0, \frac{\pi}{2}\right]$$
Since
\begin{align*}
\frac{dy}{dt}&=-\sin^2 x + \cos^2 x\\
&=\cos 2x
\end{align*}
Find the critical points
\begin{align*}
f'(t)&=0\\
\cos 2x&=0
\end{align*}
Then $x= \pi/4$. Because $\pi/4\in [0,\pi/2]$, then $f(x)$ has a critical point $\pi/4$, so \begin{align*} f(0)&=0 \\ f(\pi/2 )&= 0\\ f(\pi/4)&= 0.5 \end{align*}
Then the maximum is $f(\pi/4) =0.5$ and the minimum is $f( 0 ) =f(\pi/2)=0$
