Find the maximum value of $u$ in the following BVP
\begin{align*} u_t &= u_{xx}, \; \; t,x \in (0,1) \\ u(0,t) &= 2t^2-t \\ u(1,t) &= \sin \pi t \\ u(x,0) &= x(1-x) \\ \end{align*}
Attempt
We know maximum can only occur at the boundaries. So if we call $A = \max u(0,t) $, $B = \max u(1,t)$, $C= \max u(x,0)$, then
$$ \max u = \max (A,B,C) $$
Now, clearly, $\max u(1,t) = 1$ for any $t$ and $\max u(x,0) = 1/4$ since it is parabola opens down with vertex 1/2.
Now, for $\max u(0,t) = 2t^2 - t$, notice that it is increasing so its maximum must occur at $t=1$ thus $\max u(0,t) = 2-1 = 1$
So, we have $\max u = \max(1,1/4,1) = \boxed{1}$
Is this solution correct?
Your analysis is correct save for the computation of $\max_{0 \le t \le 1} (2t^2 - t)$. This function has a critical point in $\frac 1 4$, which turns out to be a local minimum. Since the function is decreasing on $[0, \frac 1 4]$ and increasing on $[\frac 1 4, 1]$, its maximum will be attained at the endpoints: either in $0$, or in $1$. In these points the function has the values $0$ and $1$, respectively. This means that its global maximum is $1$, which coincides with what you have found, but your argument was mistaken (the function is not increasing on $[0,1]$).
The rest of your analysis is correct: the maximum of $\sin \pi t$ is reached in $t = \frac 1 2$ and is $1$, and the maximum of $x - x^2$ is reached in $\frac 1 2$ and is $\frac 1 4$. By the weak parabolic maximum principle the maximum of $u$ will be attained on the parabolic boundary, so it will be $1$.