Chapter 5.4, Problem 32E

a

To determine

To show: that the critical numbers of C occur at the numbers

Answer to Problem 32E

critical number of c occur at the number t where $\Rightarrow \text{C(t)=(f(t)+g(t))}$

Explanation of Solution

Given information: $\text{C(t)=}\frac{1}{\text{t}}{\displaystyle {\int }_0^{\text{r}}[\text{f(s)}}+\text{g(s)]ds}$

$\text{C(t)=}\frac{1}{\text{t}}{\displaystyle {\int }_0^{\text{r}}[\text{f(s)}}+\text{g(s)]ds}$

  $\Rightarrow \text{C(t)=}\frac{1}{\text{t}}\{\text{f(t)-f(0)+g(t)-g(0)}}$

  $\Rightarrow \text{C(t)=}\frac{1}{\text{t}}\{\text{f(t)-f(0)+g(t)-g(0)}}$

  $\Rightarrow \text{C(t)=}\frac{1}{\text{t}}\{\text{f(t)+g(t)}}$

Hence , critical number of c occur at the number t where $\Rightarrow \text{C(t)=(f(t)+g(t))}$

b

To determine

the length of time T for the total depreciation .

Answer to Problem 32E

  $\text{f(t)}=\frac{29\text{v}}{15}$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{ f(t)=}\frac{\text{v}}{15}-\frac{\text{v}}{450}\text{t if 0<t}\le \text{30}\\ \text{ 0 if t>0}\\ \text{and g(t)=}\frac{{\text{vt}}^2}{12900}\text{ t>0}\end{array}$

  $\begin{array}{l}\text{f(t)=}{\displaystyle {\int }_0^{30}\frac{\text{v}}{15}}\text{dt- }{\displaystyle {\int }_0^{30}\frac{\text{v}}{30}}\text{dt}\\ \text{f(t)}=\frac{\text{v}}{15}·30-\frac{\text{v}}{450}·30\\ \text{f(t)}=2\text{v-}\frac{\text{v}}{15}\\ \text{f(t)}=\frac{29\text{v}}{15}\end{array}$

c

To determine

the length of absolute minimum .

Answer to Problem 32E

we get minimum value 0 .

Explanation of Solution

Given information: C on $(0,\text{T]}$

The absolute minimum value for maxima and minima . differentiating above equation equate to zero we get minimum value 0

d

To determine

To sketch: the graph of C.

Answer to Problem 32E

the graph of C and $\text{f+g}$ is as liner.

Explanation of Solution

Given information: $\text{f+g}$in the same coordinate.

the graph of C and $\text{f+g}$ is as liner.