Consider the triangles with vertices A 2 ,   1 ,   B 0 ,   0 and C t ,   4 ,   t =…

Given,

The triangles with vertices $A\left(2, 1\right), B\left(0, 0\right)$ and $C\left(t, 4\right), t=\left[0, 4\right]$,

And the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively,

Now to minimise $CA+CB$, in below diagram take image of $B$ in $y=4$,

We get, $B^{\text{'}}=\left(0, 8\right)$

 

Now finding, equation of $A{B}^{\text{'}}$ we get,

$y-8=\frac{-7}{2}\left(x-0\right)$

Now, putting $y=4$ in above equation we get,

$-4=\frac{-7}{2}\left(x\right)$

$\Rightarrow x=\frac{8}{7}\Rightarrow \beta =\frac{8}{7}$

Now, maximum perimeter will be possible if $\alpha =0$ or $4$

Now taking $\alpha =0$ we get,

$AB=\sqrt{5}, BC=4 \& AC=\sqrt{13}$

Now when $\alpha =4$ we get,

$AB=\sqrt{5}, BC=4\sqrt{2} \& AC=\sqrt{13}$

Now on comparing the perimeter we get, maximum perimeter at $\alpha =4$

Hence, $6\alpha +21\beta =48$