Question

the number of people in a line at a time t can be modeled by the cubic function $f(t) = -0.40t^3 + 5.44t^2 - 21.93t + 48.18$. based on the model, how many hours after the ticket window opened did the number of people in line reach a minimum before increasing? a 0 b 3 c 6 d 9

Explanation:

Step1: Find the derivative of \( f(t) \)

To find the minimum of a function, we first find its derivative. The function is \( f(t) = -0.40t^3 + 5.44t^2 - 21.93t + 48.18 \). Using the power rule, the derivative \( f'(t) \) is:
\( f'(t) = -3\times0.40t^2 + 2\times5.44t - 21.93 = -1.2t^2 + 10.88t - 21.93 \)

Step2: Solve \( f'(t) = 0 \) for critical points

We set the derivative equal to zero to find critical points: \( -1.2t^2 + 10.88t - 21.93 = 0 \). Multiply both sides by -1 to simplify: \( 1.2t^2 - 10.88t + 21.93 = 0 \). We can use the quadratic formula \( t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 1.2 \), \( b=-10.88 \), \( c = 21.93 \).

First, calculate the discriminant \( D = b^2 - 4ac = (-10.88)^2 - 4\times1.2\times21.93 \)
\( D = 118.3744 - 105.264 = 13.1104 \)

Then, \( t=\frac{10.88\pm\sqrt{13.1104}}{2\times1.2} \). \( \sqrt{13.1104}\approx3.62 \)

So, \( t_1=\frac{10.88 + 3.62}{2.4}=\frac{14.5}{2.4}\approx6.04 \) and \( t_2=\frac{10.88 - 3.62}{2.4}=\frac{7.26}{2.4}\approx3.025 \)

Step3: Determine which critical point is a minimum

We can use the second derivative test. Find the second derivative \( f''(t) = -2.4t + 10.88 \)

For \( t = 3 \) (approx \( t_2 \)): \( f''(3)= -2.4\times3 + 10.88 = -7.2 + 10.88 = 3.68>0 \), so the function is concave up at \( t = 3 \), meaning it's a local minimum.

For \( t = 6 \) (approx \( t_1 \)): \( f''(6)= -2.4\times6 + 10.88 = -14.4 + 10.88 = -3.52<0 \), so it's a local maximum.

Since we want the minimum before increasing, we take the smaller \( t \) value which is approximately 3.

Answer:

B. 3