Key Concepts and Formulas

• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Perimeter of a Triangle: The perimeter of a triangle is the sum of the lengths of its three sides.
• Finding Maxima/Minima: To find the maximum and minimum values of a function on a closed interval, we can use calculus. We find the critical points by taking the derivative, setting it to zero, and solving. We then evaluate the function at the critical points and the endpoints of the interval to determine the maximum and minimum values.

Step-by-Step Solution

Step 1: Calculate the side lengths.

We are given the vertices $A(2,1)$, $B(0,0)$, and $C(t,4)$. We need to find the lengths of the sides $AB$, $BC$, and $CA$ in terms of $t$.

• $AB = \sqrt{(2-0)^2 + (1-0)^2} = \sqrt{4 + 1} = \sqrt{5}$.
• $BC = \sqrt{(t-0)^2 + (4-0)^2} = \sqrt{t^2 + 16}$.
• $CA = \sqrt{(t-2)^2 + (4-1)^2} = \sqrt{t^2 - 4t + 4 + 9} = \sqrt{t^2 - 4t + 13}$.

Step 2: Define the perimeter function.

The perimeter $P(t)$ of the triangle is the sum of the side lengths: $P(t) = AB + BC + CA = \sqrt{5} + \sqrt{t^2 + 16} + \sqrt{t^2 - 4t + 13}.$

Step 3: Find the derivative of the perimeter function.

To find the critical points, we need to find the derivative of $P(t)$ with respect to $t$: $P'(t) = \frac{d}{dt} \left(\sqrt{5} + \sqrt{t^2 + 16} + \sqrt{t^2 - 4t + 13}\right) = 0 + \frac{2t}{2\sqrt{t^2 + 16}} + \frac{2t - 4}{2\sqrt{t^2 - 4t + 13}}.$ $P'(t) = \frac{t}{\sqrt{t^2 + 16}} + \frac{t - 2}{\sqrt{t^2 - 4t + 13}}.$

Step 4: Find the critical points.

To find the critical points, we set $P'(t) = 0$: $\frac{t}{\sqrt{t^2 + 16}} + \frac{t - 2}{\sqrt{t^2 - 4t + 13}} = 0.$ $\frac{t}{\sqrt{t^2 + 16}} = -\frac{t - 2}{\sqrt{t^2 - 4t + 13}}.$ Square both sides: $\frac{t^2}{t^2 + 16} = \frac{(t - 2)^2}{t^2 - 4t + 13}.$ $t^2(t^2 - 4t + 13) = (t^2 + 16)(t^2 - 4t + 4).$ $t^4 - 4t^3 + 13t^2 = t^4 - 4t^3 + 4t^2 + 16t^2 - 64t + 64.$ $t^4 - 4t^3 + 13t^2 = t^4 - 4t^3 + 20t^2 - 64t + 64.$ $0 = 7t^2 - 64t + 64.$ Now we use the quadratic formula to solve for $t$: $t = \frac{-(-64) \pm \sqrt{(-64)^2 - 4(7)(64)}}{2(7)} = \frac{64 \pm \sqrt{4096 - 1792}}{14} = \frac{64 \pm \sqrt{2304}}{14} = \frac{64 \pm 48}{14}.$ So, $t = \frac{64 + 48}{14} = \frac{112}{14} = 8$ or $t = \frac{64 - 48}{14} = \frac{16}{14} = \frac{8}{7}$. Since $t \in [0,4]$, the only critical point in the interval is $t = \frac{8}{7}$.

Step 5: Evaluate the perimeter function at the critical point and endpoints.

We need to evaluate $P(t)$ at $t = 0$, $t = \frac{8}{7}$, and $t = 4$.

• $P(0) = \sqrt{5} + \sqrt{16} + \sqrt{13} = \sqrt{5} + 4 + \sqrt{13} \approx 2.236 + 4 + 3.606 = 9.842$.
• $P(4) = \sqrt{5} + \sqrt{16 + 16} + \sqrt{16 - 16 + 13} = \sqrt{5} + \sqrt{32} + \sqrt{13} = \sqrt{5} + 4\sqrt{2} + \sqrt{13} \approx 2.236 + 5.657 + 3.606 = 11.499$.
• $P\left(\frac{8}{7}\right) = \sqrt{5} + \sqrt{\left(\frac{8}{7}\right)^2 + 16} + \sqrt{\left(\frac{8}{7}\right)^2 - 4\left(\frac{8}{7}\right) + 13} = \sqrt{5} + \sqrt{\frac{64}{49} + \frac{784}{49}} + \sqrt{\frac{64}{49} - \frac{224}{49} + \frac{637}{49}} = \sqrt{5} + \sqrt{\frac{848}{49}} + \sqrt{\frac{477}{49}} = \sqrt{5} + \frac{\sqrt{848}}{7} + \frac{\sqrt{477}}{7} \approx 2.236 + \frac{29.12}{7} + \frac{21.84}{7} \approx 2.236 + 4.16 + 3.12 = 9.516$.

The maximum perimeter occurs at $t = 4$, so $\alpha = 4$. The minimum perimeter occurs at $t = \frac{8}{7}$, so $\beta = \frac{8}{7}$.

Step 6: Calculate $6\alpha + 21\beta$.

$6\alpha + 21\beta = 6(4) + 21\left(\frac{8}{7}\right) = 24 + 3(8) = 24 + 24 = 48$. However, this gives us the wrong answer. Let's re-examine the endpoints and critical point. We have $P(0) \approx 9.842$, $P(8/7) \approx 9.516$, and $P(4) \approx 11.499$. Therefore, $\alpha = 4$ and $\beta = \frac{8}{7}$. $6\alpha + 21\beta = 6(4) + 21(8/7) = 24 + 24 = 48$. This is wrong.

Let's look closely at $P'(t) = \frac{t}{\sqrt{t^2 + 16}} + \frac{t - 2}{\sqrt{t^2 - 4t + 13}}$. When $t=0$, $P'(0) = 0 + \frac{-2}{\sqrt{13}} < 0$. When $t=4$, $P'(4) = \frac{4}{\sqrt{32}} + \frac{2}{\sqrt{13}} > 0$. When $t=8/7$, $P'(8/7) = 0$. So the minimum is at $t=8/7$, and the maximum is at $t=4$. There must be an error. Let's go back to $P'(t) = 0$: $\frac{t}{\sqrt{t^2+16}} = \frac{2-t}{\sqrt{t^2-4t+13}}$.

If we test $t = 1$, $\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{10}}$ which is false. If we test $t = 2$, $\frac{2}{\sqrt{20}} = 0$, which is false.

However, when we squared the equation, we introduced extraneous solutions. We must have $\frac{t}{\sqrt{t^2+16}} = \frac{2-t}{\sqrt{t^2-4t+13}}$. So $t$ must be less than 2. Since $\beta = 8/7$ minimizes the perimeter, then $\beta = 8/7$. The maximum is at $t=4$, so $\alpha = 4$. $6(4) + 21(8/7) = 24 + 24 = 48$

Upon reviewing the problem and the answer, it seems there is an error in the provided answer. Let us consider $t=0$. Then $P(0) = \sqrt{5} + 4 + \sqrt{13}$. Let us consider $t=2$. $P(2) = \sqrt{5} + \sqrt{20} + \sqrt{9} = \sqrt{5} + 2\sqrt{5} + 3 = 3\sqrt{5} + 3 \approx 9.708$.

Let's consider $t = 1$. $P(1) = \sqrt{5} + \sqrt{17} + \sqrt{10} \approx 2.236 + 4.123 + 3.162 = 9.521$. $P(8/7) \approx 9.516$ is indeed the minimum value. Then $\beta = 8/7$. Since the maximum occurs at the endpoint $t=4$, then $\alpha = 4$. $6\alpha + 21\beta = 6(4) + 21(8/7) = 24 + 24 = 48$.

There is likely an error in the question or answer. If we assume the minimum is at $t=0$ and the maximum is at $t=4$, then $\beta=0$ and $\alpha=4$. Then $6\alpha + 21\beta = 6(4) + 21(0) = 24$. If we assume the minimum is at $t=1$ and the maximum is at $t=4$, then $\beta=1$ and $\alpha=4$. Then $6\alpha + 21\beta = 6(4) + 21(1) = 24 + 21 = 45$. If the answer is 2, we need $6\alpha + 21\beta = 2$. $6\alpha = 2 - 21\beta$. $\alpha = (2 - 21\beta)/6$. If $\beta = 0$, $\alpha = 1/3$. Then $P(1/3) = \sqrt{5} + \sqrt{1/9 + 16} + \sqrt{1/9 - 4/3 + 13} = \sqrt{5} + \sqrt{145/9} + \sqrt{100/9} = \sqrt{5} + \sqrt{145}/3 + 10/3 \approx 2.236 + 4.01 + 3.333 = 9.579$.

The problem is flawed. With $\alpha = 0$ and $\beta = 1/3$, $6(0) + 21(1/3) = 7$.

Let us assume that the correct answer is $48$, then the provided answer of $2$ is wrong. If we assume the correct answer is 2, then the given information is wrong.

Common Mistakes & Tips

• Extraneous Solutions: Squaring equations can introduce extraneous solutions. Always check your solutions in the original equation.
• Endpoint Evaluation: Don't forget to evaluate the function at the endpoints of the interval. The maximum or minimum may occur at an endpoint.
• Derivative Calculation: Double-check your derivative calculations to avoid errors.

Summary

We found the perimeter function of the triangle in terms of $t$. Then, we calculated the derivative of the perimeter function and found the critical points. We evaluated the perimeter function at the critical points and the endpoints of the interval $[0,4]$. We determined that the maximum perimeter occurs at $t=4$ and the minimum perimeter occurs at $t=8/7$. Finally, we computed $6\alpha + 21\beta = 6(4) + 21(8/7) = 48$. However, the provided answer is 2. This indicates a potential issue with the question itself or the given correct answer.

Final Answer

The final answer is \boxed{48}.