Applications of Derivatives
Optimization of Perimeter in Triangle
nta_pyq_2023_apr
Grade 12
Question:
Consider the triangles with vertices $A(2,1)$, $B(0,0)$ and $C(t,4)$, $t\in[0,4]$. If the maximum and the minimum perimeters of such triangles are obtained at $t=\alpha$ and $t=\beta$ respectively, then $6\alpha+21\beta$ is equal to ___________.
Step-by-Step Solution
Key Concept: To minimise $CA+CB$, reflect $B$ in the line $y=4$ to get $B'=(0,8)$ and minimise $CA+CB=CA+CB'$ — a straight-line distance problem.
Reflect $B$ to $B'=(0,8)$. Line $AB'$: $y-8=-\frac{7}{2}x$. At $y=4$: $x=\frac{8}{7}=\beta$. For max, $t=4$ gives larger perimeter than $t=0$, so $\alpha=4$. $6(4)+21\!\left(\frac{8}{7}\right)=24+24=48$.
Correct Answer: 48