Chapter 15.3, Problem 28E

a.

To determine

Find the cost to produce one more widget when the production level is $20widgets$ .

Answer to Problem 28E

  $2dollars$

Explanation of Solution

Given information:

The Auburn Widget Corporation finds that the marginal cost function associated with producing $x$ widgets is $f\left(x\right)=80-2xdollars$ .

Calculation:

Here, marginal cost function for producing $xwidgets$ of product is:

  $f\left(x\right)=80-2x$

The marginal cost function to approximate the cost for the company to produce one more widget of product after $x$ widgets of products are produced is:

  $f\left(x\right)=0-2=-2$

Thus, the marginal cost for producing one more widget of product after producing $20widgets$ of products is:

  $f\text{'}\left(20\right)=-2$

Hence, the required answer is $2dollars$ .

b.

To determine

Find the cost if company have to double its production from $20widgetsto40widgets$ .

Answer to Problem 28E

  $400dollars$

Explanation of Solution

Given information:

The Auburn Widget Corporation finds that the marginal cost function associated with producing $x$ widgets is $f\left(x\right)=80-2xdollars$ .

Calculation:

Here, the approximate cost of the company to double its widgets from $20to40widgets$ is:

  $\begin{array}{l}{\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx\\ a=40,b=20\\ \Delta x=\frac{40-20}{n}=\frac{20}{n}\\ x_i=40+i\frac{20}{n}\end{array}$

  $\begin{array}{l}Therefore,\\ {\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=\underset{n\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i=1}^n\left(80-2x_i\right)}\frac{20}{n}\end{array}$

  $\begin{array}{l}{\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=\underset{n\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i=1}^n\left(80-2\left(40+i\frac{20}{n}\right)\right)}\frac{20}{n}\\ {\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=\underset{n\to \infty }{{\displaystyle \lim }}{\displaystyle \sum _{i=1}^n\left(80-80-\frac{40i}{n}\right)}\frac{20}{n}\end{array}$

  $\begin{array}{l}{\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=\underset{n\to \infty }{{\displaystyle \lim }}\frac{-800}{n^2}{\displaystyle {\sum }_{i=1}^n-i}\\ {\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=\underset{n\to \infty }{{\displaystyle \lim }}\frac{-800}{n^2}{\displaystyle {\sum }_{i=1}^{n}i}\end{array}$

  $\begin{array}{l}{\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=\underset{n\to \infty }{{\displaystyle \lim }}\frac{-800}{n^2}\frac{n\left(n+1\right)}{2}\\ {\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=\underset{n\to \infty }{{\displaystyle \lim }}\frac{-800}{n^2}\frac{n^2\left(1+\frac{1}{n}\right)}{2}\end{array}$

  ${\displaystyle \underset{20}{\overset{40}{\int }}\left(80-2x\right)}dx=-400$

Hence, the required answer is $400dollars$ .