Chapter 2.4, Problem 29E

a.

To determine

To find: Graph of derivative of fruit fly population. What unit should be used for horizontal and vertical axis.

Answer to Problem 29E

The graph of $P(x)$ is shown. The curve start increasing at $x=30$ and up to $x=180$ after that it becomes parallel to $x$ axis.

Explanation of Solution

Given:

The given Function for profit is:

  $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$

Where $x$ is the number of software packages sold.

Calculation: To draw the graph of above function it is required to substitute different values of $x$ and then find corresponding values of $P(x)$ .Plot the so obtained point on the graph to trace the curve.

Let’s take $x=0$ ,then find the value of $P(x)$.

  $\begin{array}{l}P(0)=\frac{10}{1+50\cdot 2^{5-\left(0.1\right)0}}\\ P(0)=\frac{10}{1+50\cdot 2^5}\\ P(0)=.00624\end{array}$

Substitute $x=10$ :

  $\begin{array}{l}P(10)=\frac{10}{1+50\cdot 2^{5-\left(0.1\right)10}}\\ P(10)=\frac{10}{1+50\cdot 2^{5-1}}\\ P(10)=\frac{10}{1+50\cdot 2^4}\\ P(10)=.01248\end{array}$

Substitute $x=20$ :

  $\begin{array}{l}P(10)=\frac{10}{1+50\cdot 2^{5-\left(0.1\right)20}}\\ P(10)=\frac{10}{1+{502}^{5-2}}\\ P(10)=\frac{10}{1+{502}^3}\\ P(10)=.02493\end{array}$

Similarly, more points can be found and trace them on graph.

The graph of $P(x)$ is given below:

  

b.

To determine

To find: What values of $x$ make sense in the graph.

Answer to Problem 29E

The values of $x=30$ and $x=180$ make sense.

Explanation of Solution

Given:

The given Function for profit is:

  $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$

Where $x$ is the number of software packages sold.

The graph of $P(x)$ is given:

  

Calculation:

The curve start increasing at $x=30$ and up to $x=180$ after that it becomes parallel to $x$ axis.

c.

To determine

To find: Draw the graph of $P\text{'}\left(x\right)$ by using NDER, for what values of $x$ is $P$ relatively sensitive to change in $x$ .

Answer to Problem 29E

The value of $P\text{'}\left(x\right)$ is maximum at $x=106.439$ after that is starts decreasing.

Explanation of Solution

Given:

The given Function for profit is:

  $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$

Where $x$ is the number of software packages sold.

Concept used: Numerical derivative is defined as:

NDER $\left(f\left(x\right),x\right)=\frac{f\left(x+0.001\right)-f\left(x-0.001\right)}{0.002}$

Calculation:

The derivative of $P\left(x\right)$ is:

  $P\text{'}\left(x\right)=\frac{34.657\cdot 2^{-0.1x+5}}{{\left(1+50\cdot 2^{-0.1x+5}\right)}^2}$

The graph of $P\text{'}\left(x\right)$ is below:

  

The value of $P\text{'}\left(x\right)$ is maximum at $x=106.439$ after that is starts decreasing.

d.

To determine

To find: What is the profit when marginal profit is greatest.

Answer to Problem 29E

The marginal profit is greatest at $x=106.439$ and profit is 5 at $x=106.439$ .

Explanation of Solution

Given:

The given Function for profit is:

  $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$

Where $x$ is the number of software packages sold.

The graph of $P\text{'}\left(x\right)$ is given below:

  

Calculation:

It is clear from the graph that marginal profit is maximum at $x=106.439$ .

Now substitute $x=106.439$ in $P\left(x\right)$ to find the profit:

  $\begin{array}{l}P(106.439)=\frac{10}{1+50\cdot 2^{5-0.1\left(106.439\right)}}\\ P(106.439)=5\end{array}$

The profit is 5 when marginal profit is greatest.

e.

To determine

To find: What is the profit when 50,100,125,150,175,300 units are sold respectively.

Answer to Problem 29E

The profit when 50,100,125,150,175,300 units are sold is 0.1960 ,3.90, 7.83 ,9.5,9.9 respectively.

Explanation of Solution

Given:

The given Function for profit is:

  $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$

Where $x$ is the number of software packages sold.

Calculation:

Substitute $x=50$ in the given function of profit:

  $\begin{array}{l}P(50)=\frac{10}{1+50\cdot 2^{5-0.1(50)}}\\ P(50)=0.1960\end{array}$

Substitute $x=100$ in the given function of profit:

  $\begin{array}{l}P(100)=\frac{10}{1+50\cdot 2^{5-0.1(100)}}\\ P(100)=3.90\end{array}$

Substitute $x=125$ in the given function of profit:

  $\begin{array}{l}P(125)=\frac{10}{1+50\cdot 2^{5-0.1(125)}}\\ P(125)=7.83\end{array}$

Substitute $x=150$ in the given function of profit:

  $\begin{array}{l}P(150)=\frac{10}{1+50\cdot 2^{5-0.1(150)}}\\ P(150)=9.5\end{array}$

Substitute $x=175$ in the given function of profit:

  $\begin{array}{l}P(175)=\frac{10}{1+50\cdot 2^{5-0.1(175)}}\\ P(175)=9.9\end{array}$

f.

To determine

To find: What is $x\to \infty P\left(x\right)$ , what is maximum profit?

Answer to Problem 29E

  $x\to \infty P\left(x\right)$ , the maximum profit is 10.

Explanation of Solution

Given:

The given Function for profit is:

  $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$

Where $x$ is the number of software packages sold.

Calculation:

  $\begin{array}{l}x\to \infty P(x)=\frac{10}{1+50\cdot 2^{5-0.1\left(\infty \right)}}\\ x\to \infty P(x)=10\end{array}$

Maximum profit is 10.

f.

To determine

To find: Is there practical reasoning that where profit is maximum explain?

Answer to Problem 29E

There is an asymptote at $y=10$ that means this is the maximum profit.

Explanation of Solution

Given: The given Function for profit is:

Where $x$ is the number of software packages sold.

Calculation:

The graph of above function is given below:

  

There is asymptote at $y=10$ this means graph attains maximum height at $y=10$ this gives the explanation that maximum profit is 10.