Q. 66 - Calculus

Question

 Len’s company produces different-sized cylindrical cans that are each 6 inches tall. The cost to produce a can with radius r is $C\left(r\right)=10{\mathrm{\pi r}}^2+24\mathrm{\pi r}$cents.

(a) Len’s boss wants him to construct the cans so that the cost of each can is within 25 cents of \(4.00. Given these cost requirements, what is the acceptable range of values for r? 

(b) Len’s boss now says that he wants the cans to cost within 10 cents of \)4.00. Under these new cost requirements, what is the acceptable range of values for r? 

(c) Interpret this problem in terms of delta and epsilon ranges. Specifically, what is c? What is L? What is epsilon for part (a) and part (b)? What are the corresponding values of delta? Illustrate these values of c, L, epsilon, and delta on a graph of C(r).

Step-by-Step Solution

Answer

Part(a) The range for the radius is $\left[2.46,2.67\right]$

Part(b) The range for the radius is $\left[2.52,2.60\right]$

Part(c)  $C\left(r\right)=100{\mathrm{\pi r}}^2+24\mathrm{\pi r}, \mathrm{c}=6, \mathrm{L}=4$

Part(a) Step 1. Given information

The given function is $C\left(r\right)=10{\mathrm{\pi r}}^2+24\mathrm{\pi r}$ and the graph is

Part(a) Step 2. Explanation

Find the value of C(r) by substituting the given values in it.

$$\begin{gathered} C\left(r\right)=4-0.25=3.75 \\ \mathrm{and} \\ C\left(r\right)=4+0.25=4.25 \end{gathered}$$

Plot the points on the graph, we get,

Hence, the range is $\left[2.46,2.67\right]$

Part(b) Step 1. Explanation

Find the value of C(r)  by substituting the given values in it.

$$\begin{gathered} C\left(r\right)=4-0.10=3.90 \\ \mathrm{And} \\ C\left(r\right)=4+0.10=4.10 \end{gathered}$$

Plot the points on the graph, we get,

Hence, the range is $\left[2.52,2.60\right]$

Part(c) Step 1. Explanation

Considering the statement from part(a), the limit expression can be made as,

$\underset{r\to 6}{lim}C\left(r\right)=4,\epsilon =0.25$

Here, we have,

$C\left(r\right)=100{\mathrm{\pi r}}^2+24\mathrm{\pi r}, \mathrm{c}=6, \mathrm{L}=4$

Again, considering the statement from part(a), the limit expression can be made as, 

$\underset{r\to 6}{lim}C\left(r\right)=4, \epsilon =0.10$

Here, we have, 

$C\left(r\right)=100{\mathrm{\pi r}}^2+24\mathrm{\pi r}, \mathrm{c}=6, \mathrm{L}=4$