JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Application of Derivatives

Given vertices: $A(2,1)$, $B(0,0)$, $C(t,4)$ where $t\in [0,4]$.

The side $AB$ is fixed with length $AB=\sqrt{(2-0{)}^2+(1-0{)}^2}=\sqrt{5}$.

Perimeter $P=AB+BC+CA=\sqrt{5}+\sqrt{t^2+16}+\sqrt{(t-2{)}^2+9}$

To minimize $BC+CA$, we use the reflection principle. Reflect $B$ across the line $y=4$ to get $B_1(0,8)$. The minimum occurs when $A$, $C$, and $B_1$ are collinear.

Slope of $A{B}_1=\frac{8-1}{0-2}=-\frac{7}{2}$

Line $A{B}_1$: $y-1=-\frac{7}{2}(x-2)$, which gives $7x+2y=16$.

At $y=4$: $7x+8=16\Rightarrow x=\frac{8}{7}$

So $\beta =\frac{8}{7}$.

For maximum perimeter, $C$ should be at the extreme of the interval, i.e., $t=4$ (farthest from both $A$ and $B$).

So $\alpha =4$.

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