Chapter 6, Problem 44RE

To determine

To calculate:the Rich’s average daily inventory and average daily holding cost.

Answer to Problem 44RE

The value is $av\left(I\right)=4800$

Explanation of Solution

Given information:

The Rich Wholesale Foods, a manufacturer of cookies, stores its cases of cookies in an air-conditioned warehouse for shipment every 14 days. Rich tries to keep 600 cases on reserve to meet occasional peaks in demand, so a typical 14-day inventory function is $I(t)=600+600t,0\le t\le 14.$

Formula used:

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

Calculation:

Calculate, Rich’s average daily inventory by using average formula

  $\begin{array}{l}av\left(I\right)=\frac{1}{14-0}{\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt}\\ av\left(I\right)=\frac{1}{14}{\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt}\end{array}$

Now, calculate the value of integral

If f is continuous at every point of [a, b], and if F is any anti-derivative of f on [a, b], then

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=F\left(b\right)-F\left(a\right)$

This part of the fundamental theorem is also called the integral evaluation theorem.

Proof

Part 1 of the fundamental theorem tells us that an anti-derivative of f exists, namely

  $G\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)dt}$

Thus, if F is any anti- derivative of f , then $F\left(x\right)=G\left(x\right)+C$ for some constant C

  $\begin{array}{l}F\left(b\right)-F\left(a\right)=\left[G\left(b\right)+C\right]-\left[G\left(a\right)+C\right]\\ =G\left(b\right)-G\left(a\right)\\ ={\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt-{\displaystyle \underset{a}{\overset{a}{\int }}f\left(t\right)}dt\\ ={\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt-0\\ F\left(b\right)-F\left(a\right)={\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt\end{array}$

So,

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt=F\left(b\right)-F\left(a\right)$

Now, use the above proof to find the value of integral.

  $f\left(t\right)=600+600t$

If F is any anti-derivative of f, then

  $f\left(t\right)=600t+300t^2$

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt=\left[600\left(14\right)+300{\left(14\right)}^2\right]}-\left[600\left(0\right)+300{\left(0\right)}^2\right]\\ =\left[67,200\right]-\left[0\right]\\ {\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt=67,200}\end{array}$

Now, calculate the value of average

  $\begin{array}{l}av\left(I\right)=\frac{1}{14}{\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt}\\ =\frac{1}{14}\times 67,200\\ av\left(I\right)=4800\end{array}$

Since, holding cost for each case is 4¢. Thus, holding cost is given by

  $c\left(t\right)=0.04I\left(t\right)$

Now, take average both sides

  $\begin{array}{l}av\left[c\left(t\right)\right]=av\left[0.04I\left(t\right)\right]\\ =0.04\times av\left(I\right)\\ =0.04\times 4800\\ av\left[c\left(t\right)\right]=192\end{array}$

This part of the fundamental theorem is also called the integral evaluation theorem.

Proof:

Part 1 of the fundamental theorem tells us that an anti-derivative or f exists, namely

  $G\left(x\right)={\displaystyle \underset{a}{\overset{x}{\int }}f\left(t\right)}dt$

Thus, if F is any anti-derivative of f , then $F\left(x\right)=G\left(x\right)+C$ for some constant C $\begin{array}{l}F\left(b\right)-F\left(a\right)=\left[G\left(b\right)+C\right]-\left[G\left(a\right)+C\right]\\ =G\left(b\right)-G\left(a\right)\\ ={\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt-{\displaystyle \underset{a}{\overset{a}{\int }}f\left(t\right)}dt\\ ={\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt-0\\ F\left(b\right)-F\left(a\right)={\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt\end{array}$

So,

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(t\right)}dt=F\left(b\right)-F\left(a\right)$

Now, use the above proof to find the value of integral.

  $f\left(t\right)=600+600t$

If F is any anti-derivative of f , then

  $f\left(t\right)=600t+300t^2$

So,

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt=\left[600\left(14\right)+300{\left(14\right)}^2\right]}-\left[600\left(0\right)+300{\left(0\right)}^2\right]\\ =\left[67,200\right]-\left[0\right]\\ {\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt=67,200}\end{array}$

Now, calculate the value of average

  $\begin{array}{l}av\left(I\right)=\frac{1}{14}{\displaystyle \underset{0}{\overset{14}{\int }}\left(600+600t\right)dt}\\ =\frac{1}{14}\times 67,200\\ av\left(I\right)=4800\end{array}$

Conclusion:

The value is $av\left(I\right)=4800$