Chapter 6.7, Problem 39E

a.

To determine

To find:The time it takes the sunflower seedlings to reach a height of 200 centimeters.

Answer to Problem 39E

It takes 5.9 weeks for the sunflower seedling to reach a height of 200 centimetres.

Explanation of Solution

Given information:

The height h of the seedlings after t weeks can be modelled by the logistic function $h\left(t\right)=\frac{256}{1+13e^{-0.65t}}$ .

Calculation :

The given logistic function is

  $h\left(t\right)=\frac{256}{1+13e^{-0.65t}}$ …… (1)

Substitute $h\left(t\right)=200$ in above equation to find the time it takes the sunflower to reach a height of 200 centimetres.

  $\begin{array}{l}200=\frac{256}{1+13e^{-0.65t}}\\ 1+13e^{-0.65t}=\frac{256}{200}\\ 13e^{-0.65t}=\frac{32}{25}-1\\ 13e^{-0.65t}=\frac{7}{25}\\ -0.65t=\ln \left(\frac{7}{325}\right)\\ -0.65t=-3.8379\\ t=\frac{-3.8379}{-0.65}\\ t=5.9\end{array}$

Hence,

It takes 5.9 weeks for the sunflower seedling to reach a height of 200 centimetres.

b.

To determine

To interpret:Use a graphing calculator to graph the function and interpret the meaning of the asymptotes in the context of this situation.

Answer to Problem 39E

The asymptotes in the context of this situation interpret that the maximum height that sunflower seedling can reach is 256.

Explanation of Solution

Given information:

The height h of the seedlings after t weeks can be modelled by the logistic function $h\left(t\right)=\frac{256}{1+13e^{-0.65t}}$ .

Calculation :

The given logistic function is

  $h\left(t\right)=\frac{256}{1+13e^{-0.65t}}$

The graph of the above function is drawn below:

  

The asymptotes in the context of this situation interpret that the maximum height that sunflower seedling can reach is 256.

Hence,

The asymptotes in the context of this situation interpret that the maximum height that sunflower seedling can reach is 256.