1. Write the perimeter as a function of $t$.

The triangle has vertices $A(2,1),\quad B(0,0),\quad C(t,4), \qquad t\in[0,4].$

Its perimeter is $P(t)=AB+BC+CA.$

Now, $AB=\sqrt{(2-0)^2+(1-0)^2}=\sqrt{5}.$

Also, $BC=\sqrt{(t-0)^2+(4-0)^2}=\sqrt{t^2+16},$ $CA=\sqrt{(t-2)^2+(4-1)^2}=\sqrt{(t-2)^2+9}.$

Hence, $P(t)=\sqrt{5}+\sqrt{t^2+16}+\sqrt{(t-2)^2+9}.$

Since $\sqrt{5}$ is constant, we only need to optimize $f(t)=\sqrt{t^2+16}+\sqrt{(t-2)^2+9}, \qquad t\in[0,4].$

2. Differentiate $f(t)$.

$f'(t)=\frac{t}{\sqrt{t^2+16}}+\frac{t-2}{\sqrt{(t-2)^2+9}}.$

For critical points, set $f'(t)=0$: $\frac{t}{\sqrt{t^2+16}}+\frac{t-2}{\sqrt{(t-2)^2+9}}=0.$

So, $\frac{t}{\sqrt{t^2+16}}=\frac{2-t}{\sqrt{(t-2)^2+9}}.$

Since on $[0,4]$, the right side is nonnegative only when $t\le 2$, the critical point must lie in $[0,2]$.

Now square both sides: $\frac{t^2}{t^2+16}=\frac{(2-t)^2}{(t-2)^2+9}.$

But $(2-t)^2=(t-2)^2$, so $\frac{t^2}{t^2+16}=\frac{(t-2)^2}{(t-2)^2+9}.$

Cross-multiplying, $t^2\big((t-2)^2+9\big)=(t-2)^2(t^2+16).$

Expand indirectly: $t^2(t-2)^2+9t^2=t^2(t-2)^2+16(t-2)^2.$

Thus, $9t^2=16(t-2)^2.$

So, $3t=\pm 4(t-2).$

Case 1: $3t=4t-8 \implies t=8.$ Not in $[0,4]$.

Case 2: $3t=-4t+8 \implies 7t=8 \implies t=\frac{8}{7}.$

Thus the only critical point in $[0,4]$ is $t=\frac{8}{7}.$

3. Determine whether this gives minimum or maximum.

Check derivative sign:

• At $t=0$, $f'(0)=0+\frac{-2}{\sqrt{13}}<0.$
• At $t=2$, $f'(2)=\frac{2}{\sqrt{20}}+0>0.$

So $f'(t)$ changes from negative to positive at $t=\frac{8}{7}$, hence $\beta=\frac{8}{7}$ is the point of minimum perimeter.

4. Find the maximum on $[0,4]$.

Since the only interior critical point gives a minimum, the maximum occurs at an endpoint: $t=0$ or $t=4$.

Compute: $f(0)=\sqrt{16}+\sqrt{13}=4+\sqrt{13},$ $f(4)=\sqrt{32}+\sqrt{13}=4\sqrt{2}+\sqrt{13}.$

Since $4\sqrt{2}>4$, we get $f(4)>f(0).$

Therefore the maximum occurs at $\alpha=4.$

5. Compute $6\alpha+21\beta$.

$6\alpha+21\beta=6(4)+21\left(\frac{8}{7}\right)=24+24=48.$

Thus the required integer is $\boxed{48}.$

6. Comparison with stored answer

Stored correct answer: $48$

Our derived answer is also $48$, so they agree.