Zu 7. Funktionen in Sachzusammenhängen In einem Ameisenhügel leben Tausende bis Millionen Ameisen zusammen. Die Anzahl der Tiere in einem neu gegründeten Ameisenbau kann durch die Funktion A(t)= 1/2 t^3-9t^2+48t angenähert werden (t ist die Zeit in Monaten und A (t) die Anzahl an Ameisen in 1000 Tieren) a) Berechnen Sie, zu welchem Zeitpunkt (0≤ t≤ 9) die Population am größten war. Bestimmen Sie die An- zahl der Tiere b) Geben Sie den Zeitraum an, in dem sich die Anzahl der Tiere verkleinerte. c) Berechnen Sie, zu welchem Zeitpunkt (0≤ t≤ 9) die Population am stärksten zunimmt und zu welchem Zeitpunkt die Population am stärksten abnimmt. d) Begründen Sie, warum die Funktion für große t zur Beschreibung der Population nicht geeignet ist.

Question 1:

1. Calculate the first derivative of A(t)
   $$A(t) = \frac{1}{2}t^{3} - 9t^{2} + 48t$$$$A(t)=\frac{1}{2}{t}^3-9t^2+48t$$
   $$A'(t) = \frac{3}{2}t^{2} - 18t + 48$$$$A^{\prime }(t)=\frac{3}{2}{t}^2-18t+48$$

2. Find the critical points by setting A'(t) = 0
   $$\frac{3}{2}t^{2} - 18t + 48 = 0$$$$\frac{3}{2}{t}^2-18t+48=0$$
   Multiply by $$\frac{2}{3}$$$$\frac{2}{3}$$:
   $$t^{2} - 12t + 32 = 0$$$$t^2-12t+32=0$$
   $$(t - 4)(t - 8) = 0$$$$(t-4)(t-8)=0$$
   $$t = 4$$$$t=4$$ or $$t = 8$$$$t=8$$

3. Calculate the second derivative of A(t)
   $$A''(t) = 3t - 18$$$$A^{\prime \prime }(t)=3t-18$$

4. Determine if the critical points are maxima or minima
   For $$t = 4$$$$t=4$$: $$A''(4) = 3(4) - 18 = 12 - 18 = -6 < 0$$$$A^{\prime \prime }(4)=3(4)-18=12-18=-6<0$$, so $$t = 4$$$$t=4$$ is a local maximum.
   For $$t = 8$$$$t=8$$: $$A''(8) = 3(8) - 18 = 24 - 18 = 6 > 0$$$$A^{\prime \prime }(8)=3(8)-18=24-18=6>0$$, so $$t = 8$$$$t=8$$ is a local minimum.

5. Evaluate A(t) at the critical points and endpoints to find the absolute maximum on [0, 9]
   $$A(0) = \frac{1}{2}(0)^{3} - 9(0)^{2} + 48(0) = 0$$$$A(0)=\frac{1}{2}(0{)}^3-9(0{)}^2+48(0)=0$$
   $$A(4) = \frac{1}{2}(4)^{3} - 9(4)^{2} + 48(4) = \frac{1}{2}(64) - 9(16) + 192 = 32 - 144 + 192 = 80$$$$A(4)=\frac{1}{2}(4{)}^3-9(4{)}^2+48(4)=\frac{1}{2}(64)-9(16)+192=32-144+192=80$$
   $$A(8) = \frac{1}{2}(8)^{3} - 9(8)^{2} + 48(8) = \frac{1}{2}(512) - 9(64) + 384 = 256 - 576 + 384 = 64$$$$A(8)=\frac{1}{2}(8{)}^3-9(8{)}^2+48(8)=\frac{1}{2}(512)-9(64)+384=256-576+384=64$$
   $$A(9) = \frac{1}{2}(9)^{3} - 9(9)^{2} + 48(9) = \frac{1}{2}(729) - 9(81) + 432 = 364.5 - 729 + 432 = 67.5$$$$A(9)=\frac{1}{2}(9{)}^3-9(9{)}^2+48(9)=\frac{1}{2}(729)-9(81)+432=364.5-729+432=67.5$$

6. Determine the time and population size when the population was largest
   The maximum population occurs at $$t = 4$$$$t=4$$ months, and the population size is $$A(4) = 80$$$$A(4)=80$$. Since A(t) is in 1000s of animals, the population is $$80 \times 1000 = 80000$$$$80\times 1000=80000$$ animals.
   The answer is: The population was largest at t = 4 months with 80,000 animals.

Question 2:

1. Identify the intervals where A'(t) < 0
   We know $$A'(t) = \frac{3}{2}t^{2} - 18t + 48 = \frac{3}{2}(t-4)(t-8)$$$$A^{\prime }(t)=\frac{3}{2}{t}^2-18t+48=\frac{3}{2}(t-4)(t-8)$$
   A'(t) < 0 when $$4 < t < 8$$$$4<t<8$$
   The answer is: The number of animals decreased in the time interval (4, 8)

Question 3:

1. Find the points of inflection by setting A''(t) = 0
   $$A''(t) = 3t - 18 = 0$$$$A^{\prime \prime }(t)=3t-18=0$$
   $$3t = 18$$$$3t=18$$
   $$t = 6$$$$t=6$$

2. Analyze the sign of A''(t) around t = 6
   For $$0 \leq t < 6$$$$0\le t<6$$, $$A''(t) < 0$$$$A^{\prime \prime }(t)<0$$, so the rate of population growth is decreasing.
   For $$6 < t \leq 9$$$$6<t\le 9$$, $$A''(t) > 0$$$$A^{\prime \prime }(t)>0$$, so the rate of population growth is increasing.

3. Determine when the population increases most rapidly
   The population increases most rapidly at $$t = 6$$$$t=6$$

4. Determine when the population decreases most rapidly
   To find when the population decreases most rapidly, we need to consider the interval where the population is decreasing, which is $$4 < t < 8$$$$4<t<8$$. Since $$A''(t) = 0$$$$A^{\prime \prime }(t)=0$$ at $$t = 6$$$$t=6$$, the rate of decrease is greatest at $$t = 6$$$$t=6$$
   The answer is: The population increases most rapidly at t = 6, and decreases most rapidly at t = 6.

Question 4:

1. Analyze the behavior of the function for large t
   As $$t$$$$t$$ becomes very large, the term $$\frac{1}{2}t^{3}$$$$\frac{1}{2}{t}^3$$ will dominate the function $$A(t) = \frac{1}{2}t^{3} - 9t^{2} + 48t$$$$A(t)=\frac{1}{2}{t}^3-9t^2+48t$$. This means that $$A(t)$$$$A(t)$$ will increase without bound as $$t$$$$t$$ increases.
2. Explain why this is not realistic for a population
   In reality, a population cannot grow indefinitely. There are limiting factors such as food supply, space, and other resources. Eventually, the population growth will slow down or even decline due to these limitations. Therefore, the function is not suitable for describing the population for large $$t$$$$t$$
   The answer is: For large t, the function A(t) increases without bound, which is not realistic for a population due to limiting factors such as food, space, and resources.