JEE Mathematics 2023 Question with Solution

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

Find: The value of $6\alpha+21\beta$, where $\alpha$ gives the maximum perimeter and $\beta$ gives the minimum perimeter.

First write the side lengths:

$$AB=\sqrt{(2-0)^2+(1-0)^2}=\sqrt{5}$$ $$BC=\sqrt{(t-0)^2+(4-0)^2}=\sqrt{t^2+16}$$ $$AC=\sqrt{(t-2)^2+(4-1)^2}=\sqrt{(t-2)^2+9}$$

So the perimeter is

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

Differentiate with respect to $t$:

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

For the minimum perimeter, set $P'(t)=0$:

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

Squaring both sides,

$$\frac{t^2}{t^2+16}=\frac{(t-2)^2}{(t-2)^2+9}$$

Cross-multiplying,

$$t^2\big((t-2)^2+9\big)=(t-2)^2(t^2+16)$$

This simplifies to

$$9t^2=16(t-2)^2$$ $$9t^2=16(t^2-4t+4)$$ $$7t^2-64t+64=0$$

Solving,

$$t=\frac{64\pm48}{14}$$

Hence,

$$t=8 \quad \text{or} \quad t=\frac{8}{7}$$

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

Now check the endpoints for the maximum perimeter:

$$P(0)=\sqrt{5}+4+\sqrt{13}$$ $$P(4)=\sqrt{5}+\sqrt{32}+\sqrt{13}$$

Since $\sqrt{32}>4$, we get $P(4)>P(0)$. Therefore the maximum perimeter occurs at $\alpha=4$.

Now compute:

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

Therefore, the required value is $48$.

Given: The perimeter depends on the moving point $C(t,4)$ on the line segment with $t\in[0,4]$.

Find: $6\alpha+21\beta$.

Since only point $C$ moves, write the perimeter as the sum of two variable distances and one constant distance:

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

The minimum must occur at an interior critical point, so solve $P'(t)=0$ to get the valid point $t=\frac{8}{7}$. Thus $\beta=\frac{8}{7}$.

For the maximum on a closed interval, it is enough to compare endpoint values. At $t=0$ and $t=4$,

$$P(0)=\sqrt{5}+4+\sqrt{13}, \qquad P(4)=\sqrt{5}+\sqrt{32}+\sqrt{13}$$

Because $\sqrt{32}>4$, the larger perimeter occurs at $t=4$. Hence $\alpha=4$.

So,

$$6\alpha+21\beta=6\cdot4+21\cdot\frac{8}{7}=48$$

Therefore, the correct answer is $48$.