Chapter 3.4, Problem 29E

a.

To determine

To graph: The function $P(x)$ .

Explanation of Solution

Given Information: Monthly profit is given by $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$ , where $x$ is the number of software packages sold.

Graph:

Putting different values of x in the function and plotting the corresponding results,

  

Thus, the function is graphed.

b.

To determine

The values of $x$ that make sense in the problem situation.

Answer to Problem 29E

The values of $x$ that make sense in the problem situation are represented by $x\ge 0$.

Explanation of Solution

Given Information: Monthly profit is given by $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$ , where $x$ is the number of software packages sold.

$x$ is the number of software packages sold, as negative number of software packages cannot be sold in a real-life situation, thus, negative values of $x$ do not make sense at all, so we are considering only those values of $x$ that are greater than 0.

Thus, $x\ge 0$ .

c.

To determine

To use: NDER to graph $P\text{'}(x)$ and then to determine the values of $x$ for which $P$ is relatively sensitive to changes in $x$ .

Answer to Problem 29E

  $P$ is relatively sensitive to changes in $x$ between $x=60andx=155$ approx.

Explanation of Solution

Given Information: Monthly profit is given by $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$ , where $x$ is the number of software packages sold.

Calculation:

  $P\text{'}(x)=\frac{0-10\cdot 50\cdot 2^5(-0.1)\cdot 2^{-0.1x}\log 2}{{(1+50\cdot 2^{5-0.1x})}^2}$(Using quotient rule and common derivative principles)

Putting different values of x in the $P\text{'}(x)$ and plotting the corresponding results,

  

Thus, as can be observed in the graph, $P$ is relatively sensitive to changes in $x$ between $x=60andx=155$ approx.

d.

To determine

The profit when marginal profit is greatest.

Answer to Problem 29E

Profit when marginal profit is greatest is $\$4924$ .

Explanation of Solution

Given Information: Monthly profit is given by $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$ , where $x$ is the number of software packages sold. Graphs from previous parts.

Graph of $P\text{'}(x)$ corresponds to marginal profit, the marginal profit is greatest at about $x=106.4$ , as $x$ should be an integer,

  $P(106)=4.924$

As profit is in thousand dollars, answer is $\$4924$ .

e.

To determine

The marginal profit when $50,100,125,150,175and300$ units are sold.

Answer to Problem 29E

The approximate values are mentioned below.

Explanation of Solution

Given Information: Monthly profit is given by $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$ , where $x$ is the number of software packages sold. Graphs from previous parts.

As graph of $P\text{'}(x)$ corresponds to marginal profit, observing the graph of $P\text{'}(x)$ from part $c$ , it can be inferred that,

  $\begin{array}{l}P\text{'}(50)=0.013\Rightarrow \$13/package\\ P\text{'}(100)=0.165\Rightarrow \$165/package\\ P\text{'}(125)=0.118\Rightarrow \$118/package\\ P\text{'}(150)=0.031\Rightarrow \$31/package\\ P\text{'}(175)=0.006\Rightarrow \$6/package\\ P\text{'}(300)={10}^{-6}\Rightarrow \$0.001/package\end{array}$

Thus, the approximate values are mentioned above.

f.

To determine

${\lim }_{x\to \infty }P(x)$ and then to state the maximum profit possible.

Answer to Problem 29E

The limit is $10$ which implies that maximum possible profit is $\$10000$ .

Explanation of Solution

Given Information: Monthly profit is given by $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$ , where $x$ is the number of software packages sold. Graphs from previous parts.

Calculation:

Calculating the limit,

  $\begin{array}{l}{\lim }_{x\to \infty }P(x)={\lim }_{x\to \infty }\frac{10}{1+50\cdot 2^{5-0.1x}}\\ \\ =\frac{10}{1+50\cdot 2^{5-0.1(\infty )}}\\ \\ =\frac{10}{1+0}\\ \\ =10\end{array}$

Thus, the limit is $10$ which implies that maximum possible profit is $\$10000$ .

g.

To determine

If there is a practical explanation to the maximum profit answer and then to explain the reasoning.

Answer to Problem 29E

Yes, because company could have reached maximum scalability.

Explanation of Solution

Given Information: Monthly profit is given by $P(x)=\frac{10}{1+50\cdot 2^{5-0.1x}}$ , where $x$ is the number of software packages sold. Graphs from previous parts.

Yes, the company may not be able to go beyond the profit level of $\$10000$ because they could have reached maximum scalability, there are no more potential clients available due to geographical limitations or some factors including increasing competition and increasing cost of workforce might be stopping them from going above this particular profit bar.